Free Math Worksheets

- Determine the possible signs of \(f(x)\) and \(g(x)\). - Can a positive number equal a negative number? - After factoring, decide whether \(x^2+1\) can equal zero over the real numbers.

1. For every real \(x\), \(x^4\ge0\), so \(f(x)\ge0\). Also, \(x^6\ge0\), so \(g(x)\le0\). The values can be equal only when both are \(0\), which occurs at \(x=0\). 2. Solve algebraically: \(2x^4=-2x^6\), so \(2x^4+2x^6=0\). 3. Factor: \(2x^4(1+x^2)=0\). 4. The factor \(2x^4\) is zero only when \(x=0\). The equation \(1+x^2=0\) has no real solution. Therefore, the only intersection is \((0, 0)\).

a) \(f(x)\) is always nonnegative and \(g(x)\) is always nonpositive, so they can be equal only when both are \(0\). b) \(2x^4(1+x^2)=0\) has the only real solution \(x=0\). The only intersection is \((0, 0)\).

- A rational function equals zero when its numerator is zero and its denominator is nonzero. - Factor out the greatest power of \(x\) from the numerator. - Does the remaining quadratic match a perfect-square trinomial? - Check whether the denominator can be zero for real \(x\).

1. The denominator \(x^2+1\) is positive for every real \(x\), so it does not exclude any real zeros. 2. Factor the numerator: \(x^4-6x^3+9x^2=x^2(x^2-6x+9)=x^2(x-3)^2\). 3. Therefore, the zeros are \(x=0\) and \(x=3\). 4. Each corresponding factor is squared, so each zero has multiplicity \(2\).

\(x=0\) has multiplicity \(2\), and \(x=3\) has multiplicity \(2\).

- Replace \(x\) with \(-x\) to test symmetry. - A product equals zero when at least one factor equals zero. - Substitute \(x=0\) for the y-intercept. - Use the degree and leading coefficient for end behavior.

1. Since \(q(-x)=q(x)\), the function is even and its graph is symmetric about the y-axis. 2. Setting each factor equal to zero gives \(x=\pm2\) and \(x=\pm5\). Each factor occurs once after factoring into linear factors, so all four zeros have multiplicity \(1\). 3. \(q(0)=\frac{1}{10}(-4)(-25)=10\), so the y-intercept is \((0, 10)\). 4. The leading term is \(0.1x^4\). Because the degree is even and the leading coefficient is positive, \(q(x)\to\infty\) as \(x\to\pm\infty\).

a) Symmetric about the y-axis b) \(x=-5, -2, 2, 5\), each with multiplicity \(1\) c) \((0, 10)\) d) \(q(x)\to\infty\) as \(x\to\pm\infty\)

- When does a product of several factors equal zero? - How does an exponent on a factor indicate the multiplicity of the corresponding zero? - For a factor such as \(x^2 + a\), consider whether a real number can have a negative square. - Ignore any nonzero constant factor when finding zeros.

1. Use the zero-product property: a product equals zero when at least one factor equals zero. 2. For a), \(x - 3 = 0\) gives \(x = 3\), which has multiplicity 1. Also, \((x + 8)^2 = 0\) gives \(x = -8\), which has multiplicity 2. 3. For b), the factor \(x^2\) gives \(x = 0\) with multiplicity 2. The factor \(x - 0.4\) gives \(x = 0.4\) with multiplicity 1. The factor \(-1\) does not affect the zeros. 4. For c), the factor \(x\) gives \(x = 0\) with multiplicity 1. The equation \(x^2 + 16 = 0\) has no real solutions, so there are no other real zeros.

a) \(x = 3\) with multiplicity 1; \(x = -8\) with multiplicity 2 b) \(x = 0\) with multiplicity 2; \(x = 0.4\) with multiplicity 1 c) \(x = 0\) with multiplicity 1; no other real zeros

- Look for a common factor in all three terms. - Use the zero-product property after factoring. - For the remaining quadratic, find two integers whose product is \(-10\) and whose sum is \(3\). - In a complete factorization, the exponent on each linear factor gives the multiplicity.

1. Set \(f(x)\) equal to zero: \(x^3 + 3x^2 - 10x = 0\). 2. Factor out \(x\): \(x(x^2 + 3x - 10) = 0\). This gives the zero \(x = 0\). 3. Factor the quadratic: \(x^2 + 3x - 10 = (x - 2)(x + 5)\). The other zeros are \(x = 2\) and \(x = -5\). 4. The complete factorization is \(f(x) = x(x - 2)(x + 5)\). Each linear factor occurs once, so every zero has multiplicity 1.

\(x = -5\), \(x = 0\), and \(x = 2\), each with multiplicity 1

- Use the number of distinct simple zeros to determine the least possible degree. - Write one linear factor for each zero. - Multiply the factors to form the polynomial. - The problem does not require expanded form unless you choose to provide it.

1. Three distinct simple zeros require a polynomial of degree at least 3. 2. The corresponding linear factors are \(x + 5\), \(x\), and \(x - 1.5\). 3. Choose a nonzero leading constant of \(1\): \(f(x) = x(x + 5)(x - 1.5)\). 4. Expanding gives \(f(x) = x^3 + 3.5x^2 - 7.5x\).

\(f(x) = x(x + 5)(x - 1.5)\), or \(f(x) = x^3 + 3.5x^2 - 7.5x\)

- The exponent on \(x - 5\) gives the multiplicity of the zero. - Compare graph behavior at zeros of even and odd multiplicity. - A sign change means the graph moves from one side of the x-axis to the other. - Compare a simple crossing with a crossing at a higher odd multiplicity.

1. At \(x = 5\), the multiplicities are 1 for \(f\), 2 for \(g\), and 3 for \(h\). 2. A zero of odd multiplicity produces a sign change, so the graph crosses the x-axis. Therefore, \(f\) crosses at \(x = 5\). 3. A zero of even multiplicity produces no sign change, so the graph touches the x-axis and turns around. Therefore, \(g\) touches at \(x = 5\). 4. Function \(h\) also changes sign because its multiplicity is odd. Since the multiplicity is 3, the graph crosses with a flattened shape and a horizontal tangent at \(x = 5\).

\(f\): crosses the x-axis and changes sign \(g\): touches the x-axis and does not change sign \(h\): crosses the x-axis with a flattened shape and changes sign

- Look for a greatest common factor in the expanded polynomial. - A product equals zero when at least one factor equals zero. - Factor or solve the remaining quadratic expression. - For an equation containing only \(x^2\) and a constant, isolate \(x^2\) and take both square roots.

1. For a), factor out \(x\): \(x(x^2 - 4x - 12) = 0\). 2. Factor the quadratic: \(x^2 - 4x - 12 = (x - 6)(x + 2)\). The real zeros are \(x = -2\), \(x = 0\), and \(x = 6\). 3. For b), use the zero-product property. The equation \(x^2 - 5 = 0\) gives \(x = \pm\sqrt{5}\), and \(x + 3 = 0\) gives \(x = -3\).

a) \(x = -2\), \(x = 0\), and \(x = 6\) b) \(x = -3\), \(x = -\sqrt{5} \approx -2.24\), and \(x = \sqrt{5} \approx 2.24\)

- A zero \(x = a\) corresponds to the factor \(x - a\). - Combine the factors for all required zeros in one product. - A general factored polynomial has the form \(a(x - x_1)(x - x_2)\cdots\), where \(a \ne 0\). - One valid counterexample is enough to disprove an impossibility claim.

1. A zero \(x = r\) corresponds to a linear factor \(x - r\). 2. The five required zeros give the factors \(x + 2\), \(x + 1\), \(x\), \(x - 1\), and \(x - 2\). 3. Their product defines the polynomial \(f(x) = x(x + 2)(x + 1)(x - 1)(x - 2)\). 4. This fifth-degree polynomial has exactly the five required zeros, so the claim is false.

The claim is false. One example is \(f(x) = x(x + 2)(x + 1)(x - 1)(x - 2)\).

- Set each function equal to zero. - Isolate the power of the variable. - Use complete factored form to determine multiplicity. - For even exponents, check whether a negative right side is possible over the real numbers.

1. For a), set the function equal to zero: \(5x^4 - 405 = 0\), so \(x^4 = 81\). The real zeros are \(x = -3\) and \(x = 3\), each with multiplicity 1. 2. For b), \(x^3 + 343 = 0\) gives \(x^3 = -343\), so \(x = -7\). This zero has multiplicity 1. 3. For c), \(x^6 = 0\) gives \(x = 0\). The factor \(x\) occurs six times, so the zero has multiplicity 6. 4. For d), \(x^4 + 16 = 0\) would require \(x^4 = -16\). An even power of a real number cannot be negative, so there are no real zeros.

a) \(x = -3\) and \(x = 3\), each with multiplicity 1 b) \(x = -7\) with multiplicity 1 c) \(x = 0\) with multiplicity 6 d) No real zeros

- Apply the difference-of-squares identity in stages. - A positive equation of the form \(x^4 = a\) has two real solutions. - Check whether the final quadratic factor has any real zeros.

1. Set \(x^4 - 81 = 0\), so \(x^4 = 81\). The real solutions are \(x = -3\) and \(x = 3\). 2. Factor the difference of squares: \(x^4 - 81 = (x^2 - 9)(x^2 + 9)\). 3. Factor \(x^2 - 9\) again: \(f(x) = (x - 3)(x + 3)(x^2 + 9)\). 4. The factor \(x^2 + 9\) has no real zeros. The linear factors each occur once, so \(x = -3\) and \(x = 3\) each have multiplicity 1.

1. \(x = -3\), \(x = 3\) 2. \(f(x) = (x - 3)(x + 3)(x^2 + 9)\); each real zero has multiplicity 1.

- First look for a greatest common factor in all terms. - Check whether the remaining trinomial is a perfect-square trinomial. - In factored form, the exponent on a linear factor gives the multiplicity of its zero.

1. Factor out the greatest common factor: \(f(x) = 2x^2(x^2 - 6x + 9)\). 2. The trinomial is a perfect square: \(x^2 - 6x + 9 = (x - 3)^2\). 3. Therefore, the complete factored form is \(f(x) = 2x^2(x - 3)^2\). 4. Set each variable factor equal to zero. The factors give \(x = 0\) and \(x = 3\). 5. Each linear factor is squared, so both zeros have multiplicity 2.

a) \(f(x) = 2x^2(x - 3)^2\) b) \(x = 0\) with multiplicity 2; \(x = 3\) with multiplicity 2

- What function value corresponds to the word “zero”? - When is a product of factors equal to zero? - Factor the quadratic expression before solving.

1. A value \(x_0\) is a zero of \(f\) exactly when the function value at that input is zero: \(f(x_0) = 0\). 2. Set \(g(x) = 0\): \((x + 5)(x^2 - 4x - 12) = 0\). 3. Factor the quadratic: \(x^2 - 4x - 12 = (x - 6)(x + 2)\). 4. Apply the zero-product property to \((x + 5)(x - 6)(x + 2) = 0\). The zeros are \(x = -5\), \(x = -2\), and \(x = 6\).

a) A value \(x_0\) is a zero of \(f\) when \(f(x_0) = 0\). b) The zeros are \(x = -5\), \(x = -2\), and \(x = 6\).

- A polynomial cannot have more real zeros than its degree. - Terms may not be written in descending order, so identify the greatest exponent carefully. - For a product of nonzero polynomials, add the degrees of the factors. - Simplify an expression before deciding its degree because leading terms may cancel.

1. In a), the highest-degree term is \(-12x^5\), so the degree is \(5\). A degree-\(5\) polynomial can have at most \(5\) real zeros. 2. In b), the degrees of the factors add: \(2 + 1 + 1 = 4\). Therefore, the degree is \(4\), and the function can have at most \(4\) real zeros. 3. In c), simplify first: \(x^2(x^3 + 4) - x^5 + 2x = x^5 + 4x^2 - x^5 + 2x = 4x^2 + 2x\). The degree is \(2\), so the function can have at most \(2\) real zeros.

a) Degree \(5\); at most \(5\) real zeros b) Degree \(4\); at most \(4\) real zeros c) Degree \(2\); at most \(2\) real zeros

- A product is zero when at least one factor is zero. - Solve the two quadratic equations separately. - Look for a perfect-square trinomial in the second factor. - The exponent on a linear factor gives the multiplicity.

1. Apply the zero-product property: \(x^2 - 5 = 0\) or \(x^2 + 4x + 4 = 0\). 2. From \(x^2 - 5 = 0\), \(x^2 = 5\), so \(x = -\sqrt{5}\) and \(x = \sqrt{5}\). Each corresponding linear factor occurs once, so each zero has multiplicity 1. 3. The second quadratic is a perfect square: \(x^2 + 4x + 4 = (x + 2)^2\). 4. Therefore, \(x = -2\) is a zero with multiplicity 2.

\(x = -\sqrt{5}\) with multiplicity 1; \(x = \sqrt{5}\) with multiplicity 1; \(x = -2\) with multiplicity 2

- What y-coordinate does every point on the x-axis have? - Look for a greatest common factor in all terms. - Factor or solve the quadratic expression that remains. - Report ordered pairs, not only x-values.

1. At an x-intercept, \(g(x) = 0\): \(\frac{1}{3}x^3 - \frac{4}{3}x^2 - 4x = 0\). 2. Factor out the greatest common factor: \(\frac{1}{3}x(x^2 - 4x - 12) = 0\). 3. Factor the quadratic: \(x^2 - 4x - 12 = (x - 6)(x + 2)\). 4. The zeros are \(x = -2\), \(x = 0\), and \(x = 6\). 5. Every x-intercept has y-coordinate \(0\), so the intercepts are \((-2, 0)\), \((0, 0)\), and \((6, 0)\).

The x-intercepts are \((-2, 0)\), \((0, 0)\), and \((6, 0)\).

- Set each linear factor equal to zero. - The exponent on a factor gives the multiplicity of its zero. - Compare the sign behavior of even and odd powers. - A polynomial changes sign at a zero of odd multiplicity.

1. For \(f\), the zeros are \(x = 3\) with multiplicity 2 and \(x = -1\) with multiplicity 1. The function does not change sign at \(x = 3\) because the multiplicity is even, and it changes sign at \(x = -1\) because the multiplicity is odd. 2. For \(g\), the zeros are \(x = 0\) with multiplicity 3 and \(x = 0.5\) with multiplicity 1. Both multiplicities are odd, so the function changes sign at both zeros. 3. For \(h\), the zeros are \(x = -\frac{1}{4}\) with multiplicity 2 and \(x = 2\) with multiplicity 3. The function does not change sign at \(x = -\frac{1}{4}\), and it changes sign at \(x = 2\).

a) \(x = 3\), multiplicity 2, no sign change; \(x = -1\), multiplicity 1, sign change b) \(x = 0\), multiplicity 3, sign change; \(x = 0.5\), multiplicity 1, sign change c) \(x = -\frac{1}{4}\), multiplicity 2, no sign change; \(x = 2\), multiplicity 3, sign change

- Identify where the graph crosses or touches the x-axis. - A crossing indicates an odd multiplicity, while a touch indicates an even multiplicity. - Use the graph's right-end behavior to determine the sign of the leading coefficient. - Check the signs inside the factors against the locations of the zeros.

1. The graph crosses the x-axis at \(x = -1\), so \(x = -1\) is a zero of odd multiplicity. It touches the x-axis at \(x = 3\), so \(x = 3\) is a zero of even multiplicity. 2. Choices b) and c) have a factor \((x + 1)\) and a squared factor \((x - 3)^2\), so both have the correct zeros and multiplicities. 3. The graph rises to the right, so the leading coefficient is positive. Choice b) has a positive leading coefficient, while choice c) has a negative leading coefficient. 4. Therefore, choice b) matches the graph.

b) \(f(x) = (x + 1)(x - 3)^2\)

- Factor out the greatest common factor from all terms. - Apply the zero-product property. - Compare the graph's behavior at a double zero with its behavior at a simple zero.

1. Set \(g(x) = 0\) and factor: \(x^3 - 3x^2 = x^2(x - 3)\). 2. By the zero-product property, \(x^2 = 0\) or \(x - 3 = 0\). Therefore, \(x = 0\) has multiplicity 2 and \(x = 3\) has multiplicity 1. 3. At the double zero \(x = 0\), the graph touches the x-axis. At the simple zero \(x = 3\), the graph crosses the x-axis.

The zeros are \(x = 0\) with multiplicity 2 and \(x = 3\) with multiplicity 1. The graph touches the x-axis at \(x = 0\) and crosses it at \(x = 3\).

- Set each factor equal to zero to identify the zeros. - Use each factor's exponent to decide whether the graph crosses or touches the x-axis. - Check the direction of the graph's right end.

1. The zeros are \(x = -1\) and \(x = 3\). 2. The factor \((x + 1)^2\) gives a double zero at \(x = -1\), so the graph touches the x-axis there. 3. The factor \((x - 3)\) gives a simple zero at \(x = 3\), so the graph crosses the x-axis there. 4. The polynomial has odd degree and a positive leading coefficient, so its right end rises. 5. Only Graph 1 has all of these features.

Graph 1

- Match the x-intercepts to the factors in each rule. - Decide whether the graph crosses or touches the x-axis at each zero. - Use the end behavior to determine the sign of the leading coefficient.

1. The graph crosses the x-axis at \(x=-3, -1, 1, 3\). Each crossing represents a zero with odd multiplicity, and the choices show each as a simple zero when present. 2. Both ends of the graph rise, so the polynomial has a positive leading coefficient. 3. Choice (1) has all four observed zeros and a positive leading coefficient, so it matches. 4. Choice (2) has the correct zeros but a negative leading coefficient, which would make both ends fall. 5. Choice (3) has a double zero at \(x=-3\), where its graph would touch rather than cross the x-axis, and it has no zero at \(x=-1\).

Choice (1), \(f(x)=0.1(x+3)(x+1)(x-1)(x-3)\), matches the graph. Choice (2) has the wrong end behavior, and choice (3) has the wrong zeros and multiplicities.

- Factorization already shows both zeros. - One zero is fixed and has even multiplicity. - The other x-intercept equals the parameter.

1. The zeros are \(x=0\), with multiplicity \(2\), and \(x=k\). 2. Read the second x-intercept from each graph. Graph p crosses at \(x=2\), so \(k=2\). Graph q crosses at \(x=-3\), so \(k=-3\). Graph r crosses at \(x=4\), so \(k=4\).

p) \(k=2\) q) \(k=-3\) r) \(k=4\)

- Use the given point to solve for \(a\). - After finding \(a\), set the two function expressions equal. - Even powers produce y-axis symmetry, which may help you anticipate another point.

1. Substitute \(P(2, 4)\) into \(f\): \(4=a(2^4)=16a\), so \(a=\frac{1}{4}\). 2. Set the functions equal: \(\frac{1}{4}x^4=x^2\). 3. Factor: \(\frac{1}{4}x^4-x^2=x^2\left(\frac{1}{4}x^2-1\right)=0\). 4. The solutions are \(x=0\) and \(x=\pm2\). 5. The corresponding points are \((0, 0)\), \((2, 4)\), and \((-2, 4)\). Excluding the given point, the other intersections are \((0, 0)\) and \((-2, 4)\).

a) \(a=\frac{1}{4}\) b) The other intersection points are \((0, 0)\) and \((-2, 4)\).

- Read zeros and multiplicities directly from the factors. - Substitute \(x=0\) to find the y-intercept. - Use the degree and leading coefficient for end behavior. - Symmetry requires corresponding zeros at opposite \(x\)-values.

1. From the factored form, \(x=-3\) is a zero of multiplicity \(1\), and \(x=1\) is a zero of multiplicity \(2\). 2. The y-intercept is \(p(0)=-0.1(3)(1)=-0.3\), so the point is \((0, -0.3)\). 3. The leading term is \(-0.1x^3\). Therefore, \(p(x)\to-\infty\) as \(x\to\infty\), and \(p(x)\to\infty\) as \(x\to-\infty\). 4. Either listed symmetry would require zeros to occur in opposite pairs with matching multiplicities. Since \(x=1\) is a double zero but \(x=-1\) is not a zero, neither symmetry is possible.

a) \(x=-3\), multiplicity \(1\); \(x=1\), multiplicity \(2\) b) \((0, -0.3)\) c) \(p(x)\to-\infty\) as \(x\to\infty\), and \(p(x)\to\infty\) as \(x\to-\infty\) d) The zeros are not paired across the origin with matching multiplicities, so neither symmetry applies.

- First look for a greatest common factor. - Use the zero-product property when an expression is written as a product. - For an expression involving only \(x^4\), \(x^2\), and a constant, try substituting \(u=x^2\). - Some factors may have no real zeros.

1. Factor \(f\): \(5x^3-20x^2+15x=5x(x^2-4x+3)=5x(x-1)(x-3)\). The real zeros are \(x=0\), \(x=1\), and \(x=3\). 2. For \(g\), the zero-product property gives \(x^2-2=0\) or \(x^2+9=0\). The first equation gives \(x=\pm\sqrt{2}\); the second has no real solutions. Thus, the real zeros are \(x=-\sqrt{2}\) and \(x=\sqrt{2}\). 3. For \(h\), let \(u=x^2\). Then \(u^2-13u+36=0\), which factors as \((u-9)(u-4)=0\). Thus, \(x^2=9\) or \(x^2=4\), giving \(x=-3, -2, 2, 3\).

a) \(x=0, 1, 3\) b) \(x=-\sqrt{2}, \sqrt{2}\) c) \(x=-3, -2, 2, 3\)

- A squared factor produces a repeated zero. - For a polynomial involving only \(x^4\), \(x^2\), and a constant, substitute \(u=x^2\). - Use the degree to check the maximum possible number of zeros, counting multiplicity. - Factor out a common power of \(x\) when every term contains one.

1. For \(p\), the zero-product property gives \(x=1.5\) and \(x=-\frac{1}{3}\). The zero \(-\frac{1}{3}\) has multiplicity \(2\). The factor \(x^2+5\) has no real zeros. 2. For \(q\), let \(u=x^2\). Then \(0.5u^2-3.5u-4=0\). Multiplying by \(2\) gives \(u^2-7u-8=(u-8)(u+1)=0\). Since \(u=x^2\ge 0\), only \(u=8\) gives real values of \(x\). Therefore, \(x=\pm2\sqrt{2}\). 3. For \(r\), factor: \(2x^4+4x^3-6x^2=2x^2(x^2+2x-3)=2x^2(x+3)(x-1)\). The real zeros are \(x=-3\), \(x=0\), and \(x=1\), with \(x=0\) having multiplicity \(2\).

a) \(x=1.5\); \(x=-\frac{1}{3}\) with multiplicity \(2\) b) \(x=-2\sqrt{2}, 2\sqrt{2}\) c) \(x=-3, 0, 1\); \(x=0\) has multiplicity \(2\)

- Look for a greatest common factor that can be factored from every term. - Check whether the remaining quadratic is a perfect-square trinomial or can be factored another way. - In factored form, the exponent on a factor gives the multiplicity of its corresponding zero.

1. Set the function equal to zero: \(x^3 - 6x^2 + 9x = 0\). 2. Factor out \(x\): \(x(x^2 - 6x + 9) = 0\). This gives the zero \(x = 0\). 3. Factor the perfect-square trinomial: \(x^2 - 6x + 9 = (x - 3)^2\). Therefore, the other zero is \(x = 3\). 4. In \(x(x - 3)^2\), the factor \(x\) occurs once, so \(x = 0\) has multiplicity 1. The factor \((x - 3)\) occurs twice, so \(x = 3\) has multiplicity 2. 5. The factored form is \(f(x) = x(x - 3)^2\).

Zeros: \(x = 0\) with multiplicity 1; \(x = 3\) with multiplicity 2 Factored form: \(f(x) = x(x - 3)^2\)

- Factor expressions of the form \(x^2-a^2\) as a difference of squares. - Set each factor equal to zero. - Check whether the same zero occurs in more than one factor. - A nonzero constant factor does not affect the zeros.

1. For part a, \(x^2-36=(x-6)(x+6)\), so \(f(x)=2(x-6)(x+6)^2\). The real zeros are \(x=6\) and \(x=-6\), with \(-6\) having multiplicity \(2\). 2. For part b, the factors give zeros \(x=0\), \(x=1.5\), and \(x=-11\). 3. For part c, \(x^2-0.25=(x-0.5)(x+0.5)\). The real zeros are \(x=-0.5\), \(x=0\), and \(x=0.5\), with \(0\) having multiplicity \(2\).

a) \(x=-6, 6\); \(-6\) has multiplicity \(2\). b) \(x=-11, 0, 1.5\) c) \(x=-0.5, 0, 0.5\); \(0\) has multiplicity \(2\).

- Begin by setting the function equal to zero. - Check whether every term has a common factor involving \(x\). - Use the zero-product property after factoring. - Find two numbers whose product is \(-16\) and whose sum is \(-6\).

1. Set \(f(x)\) equal to zero and factor out \(\frac{1}{2}x\): \(\frac{1}{2}x(x^2 - 6x - 16) = 0\). The factor \(x\) gives the zero \(x = 0\). 2. Factor the quadratic: \(x^2 - 6x - 16 = (x - 8)(x + 2)\). Therefore, the remaining zeros are \(x = 8\) and \(x = -2\). 3. The factored form is \(f(x) = \frac{1}{2}x(x - 8)(x + 2)\).

1. \(x = -2\), \(x = 0\), and \(x = 8\) 2. \(f(x) = \frac{1}{2}x(x - 8)(x + 2)\)

- Look for the greatest power of \(x\) that is common to every term. - Check whether the remaining trinomial is a perfect square. - In factored form, the exponent on a factor gives the multiplicity of its corresponding zero.

1. Set the function equal to zero and factor out \(-\frac{1}{3}x^2\): \(-\frac{1}{3}x^2(x^2 - 6x + 9) = 0\). 2. The factor \(x^2\) gives the zero \(x = 0\) with multiplicity 2. 3. Factor the perfect-square trinomial: \(x^2 - 6x + 9 = (x - 3)^2\). The factor \((x - 3)^2\) gives the zero \(x = 3\) with multiplicity 2.

\(x = 0\) with multiplicity 2; \(x = 3\) with multiplicity 2

- For the fourth-degree expression, try substituting a new variable for \(x^2\). - Look for a greatest common factor in every term. - Check whether the remaining quadratic is a perfect-square trinomial. - Use the zero-product property after factoring. - When you substitute back, check whether each equation has real solutions.

1. For a), substitute \(u = x^2\). Then \(u^2 - 3u - 4 = 0\), which factors as \((u - 4)(u + 1) = 0\). 2. The values are \(u = 4\) and \(u = -1\). Substituting back gives \(x^2 = 4\) or \(x^2 = -1\). Therefore, the real zeros are \(x = -2\) and \(x = 2\); \(x^2 = -1\) has no real solutions. 3. For b), factor out \(x\): \(g(x) = x(x^2 + 4x + 4) = x(x + 2)^2\). 4. By the zero-product property, the real zeros are \(x = 0\) and \(x = -2\).

a) \(x = -2\), \(x = 2\) b) \(x = -2\), \(x = 0\)

- At an intersection point, the two functions have the same output for the same input. - Move all terms to one side so the equation equals zero. - Look for a common factor before factoring the remaining quadratic. - After finding each x-coordinate, substitute it into either original function to find the y-coordinate.

1. At an intersection point, \(f(x) = g(x)\). Set the expressions equal: \(x^3 - 4x^2 + 5x + 1 = x + 1\). 2. Move all terms to one side: \(x^3 - 4x^2 + 4x = 0\). 3. Factor: \(x(x^2 - 4x + 4) = x(x - 2)^2 = 0\). Thus, the x-coordinates are \(x = 0\) and \(x = 2\). 4. Substitute into \(g(x) = x + 1\): \(g(0) = 1\) and \(g(2) = 3\). 5. The intersection points are \((0, 1)\) and \((2, 3)\).

\((0, 1)\) and \((2, 3)\)

- At an intersection point, the two functions have equal outputs. - Rewrite the equation so that one side is zero. - Look for a common factor in every term. - Use the zero-product property after factoring. - Substitute each x-coordinate into either original function to find the y-coordinate.

1. Set the functions equal: \(\frac{1}{2}x^3 - 2x = \frac{3}{2}x^2\). 2. Move all terms to one side and factor out \(x\): \(\frac{1}{2}x^3 - \frac{3}{2}x^2 - 2x = x\left(\frac{1}{2}x^2 - \frac{3}{2}x - 2\right) = 0\). 3. The factor \(x\) gives \(x = 0\). Multiply the quadratic equation by \(2\) to get \(x^2 - 3x - 4 = 0\), which factors as \((x - 4)(x + 1) = 0\). Thus, the other x-coordinates are \(x = 4\) and \(x = -1\). 4. Substitute into \(g(x) = \frac{3}{2}x^2\): \(g(0) = 0\), \(g(4) = 24\), and \(g(-1) = \frac{3}{2}\). 5. The intersection points are \((-1, 1.5)\), \((0, 0)\), and \((4, 24)\).

\((-1, 1.5)\), \((0, 0)\), and \((4, 24)\)

- Factor out the greatest common factor first. - Use the difference-of-squares pattern to factor the remaining quadratic. - The exponent on a linear factor gives the multiplicity of its zero. - Relate even and odd multiplicities to whether the function changes sign.

1. Factor out \(-\frac{1}{4}x^2\): \(g(x) = -\frac{1}{4}x^2(x^2 - 4)\). 2. Factor the difference of squares: \(g(x) = -\frac{1}{4}x^2(x - 2)(x + 2)\). The zeros are \(x = -2\), \(x = 0\), and \(x = 2\). 3. The factor \(x^2\) gives \(x = 0\) multiplicity 2. The factors \(x - 2\) and \(x + 2\) each occur once, so \(x = 2\) and \(x = -2\) each have multiplicity 1. 4. A graph crosses the x-axis at a zero of odd multiplicity and touches without crossing at a zero of even multiplicity. Therefore, the graph touches at \(x = 0\) and crosses at \(x = -2\) and \(x = 2\).

1. \(x = -2\), \(x = 0\), and \(x = 2\) 2. \(x = -2\): multiplicity 1; \(x = 0\): multiplicity 2; \(x = 2\): multiplicity 1 3. The graph touches the x-axis at \(x = 0\) and crosses it at \(x = -2\) and \(x = 2\).

- In factored form, add the degrees of the factors to find the degree of the polynomial. - Each factor equals zero at its corresponding zero. - The exponent on a factor gives the multiplicity of that zero. - Compare the graph behavior at zeros of even and odd multiplicity.

1. The degree is the sum of the degrees of the factors: \(2 + 1 + 1 = 4\). 2. The factor \((x + 1)^2\) gives \(x = -1\) with multiplicity 2. The factor \(x - 3\) gives \(x = 3\) with multiplicity 1. The factor \(x\) gives \(x = 0\) with multiplicity 1. 3. A polynomial graph touches the x-axis without crossing at a zero of even multiplicity. Therefore, the graph touches the x-axis at \(x = -1\).

a) \(4\) b) \(x = -1\) with multiplicity 2; \(x = 0\) with multiplicity 1; \(x = 3\) with multiplicity 1 c) The graph touches the x-axis at \(x = -1\) because that zero has even multiplicity.

- Set each factor equal to zero to identify the zero set. - Watch for additional factors such as \(x\). - A repeated linear factor creates a zero with multiplicity greater than 1. - A nonzero constant factor changes the vertical scale but not the zeros.

1. For \(f\), the factors give the zero set \(\{-2, 2, 5\}\). 2. For \(g\), factor \(x^2 - 4 = (x - 2)(x + 2)\). Its zero set is \(\{-5, -2, 2\}\). 3. For \(h\), the zero set is \(\{-2, 2, 5\}\). The factor \((x - 2)^2\) shows that \(x = 2\) is a double zero. 4. For \(i\), the additional factor \(x\) gives the zero set \(\{-2, 0, 2, 5\}\). 5. For \(j\), the nonzero factor \(\frac{1}{2}\) does not affect the zeros, so its zero set is \(\{-2, 2, 5\}\). 6. Therefore, \(f\), \(h\), and \(j\) have exactly the given zero set. Function \(h\) has a double zero at \(x = 2\), and \(i\) differs from \(f\) because \(i\) also has the zero \(x = 0\).

a) \(f\), \(h\), and \(j\) b) \(h\) has a double zero at \(x = 2\). c) The factor \(x\) gives \(i\) the additional zero \(x = 0\).

- Factor \(x^2 - 9\) as a difference of squares. - Compare the zero sets carefully, especially the signs of the third zeros. - In complete factored form, the exponent on each linear factor gives its zero’s multiplicity.

1. Function \(k\) has zeros \(x = -3\), \(x = 3\), and \(x = 1\). 2. For \(l\), factor the difference of squares: \(l(x) = (x - 3)(x + 3)(x - 3) = (x + 3)(x - 3)^2\). Its zeros are \(x = -3\) and \(x = 3\). 3. Function \(m\) also has exactly the zeros \(x = -3\) and \(x = 3\). 4. Function \(n\) has zeros \(x = -3\), \(x = 3\), and \(x = -1\). 5. Therefore, \(l\) and \(m\) answer part a), and \(k\) answers part b). In \(l(x) = (x + 3)(x - 3)^2\), \(x = -3\) has multiplicity 1 and \(x = 3\) has multiplicity 2.

a) \(l\) and \(m\) b) \(k\) c) \(x = -3\) with multiplicity 1; \(x = 3\) with multiplicity 2

- Write one linear factor for each given zero. - Use an exponent of 2 for the double zero. - Substitute the coordinates of the given point to determine the leading constant. - Check that the multiplicities add to the stated degree.

1. Use the zeros and multiplicities to write \(f(x) = ax(x + 1)^2(x - 4)\). 2. Substitute the point \((2, 18)\): \(18 = a(2)(2 + 1)^2(2 - 4)\). 3. Simplify: \(18 = a(2)(9)(-2) = -36a\), so \(a = -\frac{1}{2}\). 4. Therefore, a possible equation is \(f(x) = -\frac{1}{2}x(x + 1)^2(x - 4)\).

\(f(x) = -\frac{1}{2}x(x + 1)^2(x - 4)\)

- Crossing and touching the x-axis correspond to different zero multiplicities. - Translate each x-intercept into a linear factor. - Use the y-intercept to determine the leading constant. - Add the required multiplicities to determine the least possible degree.

1. Crossing the x-axis at \(x = -3\) and \(x = 5\) requires simple factors \(x + 3\) and \(x - 5\). Touching at \(x = 1\) requires the repeated factor \((x - 1)^2\). The least possible degree is \(1 + 1 + 2 = 4\). 2. Write \(g(x) = a(x + 3)(x - 5)(x - 1)^2\). 3. Use the y-intercept: \(-3 = a(0 + 3)(0 - 5)(0 - 1)^2 = -15a\). 4. Thus, \(a = \frac{1}{5}\), so \(g(x) = \frac{1}{5}(x + 3)(x - 5)(x - 1)^2\).

\(g(x) = \frac{1}{5}(x + 3)(x - 5)(x - 1)^2\)

- Write a linear factor for each zero, including zeros containing radicals. - Multiply the conjugate factors using the difference-of-squares pattern. - A simple zero corresponds to a factor with exponent 1.

1. Three distinct simple zeros require a polynomial of degree 3. 2. The corresponding linear factors are \(x + \sqrt{6}\), \(x - 2\), and \(x - \sqrt{6}\). 3. Choose a leading constant of \(1\): \(g(x) = (x + \sqrt{6})(x - \sqrt{6})(x - 2)\). 4. Use the difference of squares: \((x + \sqrt{6})(x - \sqrt{6}) = x^2 - 6\). 5. Therefore, \(g(x) = (x^2 - 6)(x - 2) = x^3 - 2x^2 - 6x + 12\).

\(g(x) = (x^2 - 6)(x - 2)\), or \(g(x) = x^3 - 2x^2 - 6x + 12\)

- Write a linear factor for each zero. - Use an exponent of 2 for the double zero. - Substitute the coordinates of the given point to determine the leading constant.

1. The zeros and multiplicities give \(f(x) = a(x + 2)^2(x - 3)\). 2. Substitute the point \((1, -18)\): \(-18 = a(1 + 2)^2(1 - 3)\). 3. Simplify: \(-18 = a(9)(-2) = -18a\), so \(a = 1\). 4. Therefore, \(f(x) = (x + 2)^2(x - 3)\).

\(f(x) = (x + 2)^2(x - 3)\)

- Relate the degree of a product to the number of its linear factors. - Consider what happens to degrees when factors are multiplied. - Use the zero-product property. - Repeated zeros allow fewer distinct zeros without lowering the degree. - A quadratic factor with no real zeros can also limit the number of real zeros.

1. Each distinct real zero \(x = r\) produces a distinct linear factor \(x - r\). 2. Four distinct real zeros would require at least four distinct linear factors, whose product has degree at least 4. Therefore, a third-degree polynomial can have at most three distinct real zeros. 3. For exactly one distinct real zero, \(f(x) = (x - 1)^3\) works. 4. For exactly two distinct real zeros, \(f(x) = (x - 1)^2(x - 2)\) works. 5. For exactly three distinct real zeros, \(f(x) = (x - 1)(x - 2)(x - 3)\) works.

a) Each distinct real zero gives a distinct linear factor. Four distinct zeros would require degree at least 4, so a cubic can have at most three. b) 1. \(f(x) = (x - 1)^3\) 2. \(f(x) = (x - 1)^2(x - 2)\) 3. \(f(x) = (x - 1)(x - 2)(x - 3)\)

- Set each factor equal to zero. - Read each multiplicity from the exponent on its factor. - Even multiplicity means no sign change; odd multiplicity means a sign change. - Compare multiplicity 1 with a higher odd multiplicity such as 3.

1. The factors give the zeros \(x = -3\), \(x = 0\), and \(x = 2\). 2. The factor \((x + 3)^2\) gives \(x = -3\) multiplicity 2. Because the multiplicity is even, the graph touches the x-axis and turns around. 3. The factor \(x\) gives \(x = 0\) multiplicity 1. The graph crosses the x-axis normally and changes sign. 4. The factor \((x - 2)^3\) gives \(x = 2\) multiplicity 3. The graph crosses the x-axis, changes sign, and has a flattened shape with a horizontal tangent.

\(x = -3\), multiplicity 2: touch-and-turn point \(x = 0\), multiplicity 1: regular crossing \(x = 2\), multiplicity 3: flattened crossing

- Factor out the greatest power of \(x\) common to every term. - Use the zero-product property after factoring. - Check whether the remaining quadratic is a perfect-square trinomial. - Read multiplicities from the exponents in the complete factorization.

1. Factor out the greatest common factor: \(g(x) = x^3(x^2 - 6x + 9)\). 2. Factor the perfect-square trinomial: \(x^2 - 6x + 9 = (x - 3)^2\). 3. Therefore, \(g(x) = x^3(x - 3)^2\). 4. The zero \(x = 0\) has multiplicity 3, and the zero \(x = 3\) has multiplicity 2.

Zeros: \(x = 0\) with multiplicity 3; \(x = 3\) with multiplicity 2 Factored form: \(g(x) = x^3(x - 3)^2\)

- For an expression containing \(x^4\) and \(x^2\), substitute a new variable for \(x^2\). - An exponent on a factor gives the multiplicity of its zero. - Read zeros directly from a complete factorization. - After solving for the substituted variable, convert each result back to values of \(x\).

1. For a), substitute \(u = x^2\). Then \(u^2 - 17u + 16 = 0\), which factors as \((u - 16)(u - 1) = 0\). 2. Substituting back gives \(x^2 = 16\) or \(x^2 = 1\). Thus, the zeros are \(x = -4\), \(x = -1\), \(x = 1\), and \(x = 4\), each with multiplicity 1. 3. For b), read the zeros from the factors. The factor \(x\) gives \(x = 0\) with multiplicity 1, \((x - 2)^2\) gives \(x = 2\) with multiplicity 2, and \(x + 5\) gives \(x = -5\) with multiplicity 1.

a) \(x = -4\), \(x = -1\), \(x = 1\), and \(x = 4\), each with multiplicity 1 b) \(x = -5\) with multiplicity 1; \(x = 0\) with multiplicity 1; \(x = 2\) with multiplicity 2

- Substitute a new variable for \(x^2\). - Consider the possible values of \(x^2\), \(x^4\), and other even powers for real \(x\). - Determine what happens when nonnegative terms are added to a positive constant.

1. For a), substitute \(u = x^2\). The equation \(f(x) = 0\) becomes \(u^2 + 3u + 2 = 0\). 2. Factor: \((u + 1)(u + 2) = 0\), so \(u = -1\) or \(u = -2\). 3. Substituting back gives \(x^2 = -1\) or \(x^2 = -2\). Neither equation has a real solution, so \(f\) has no real zeros. 4. For b), every even power of a real number is nonnegative. Multiplying those powers by positive coefficients keeps each variable term nonnegative. Adding a positive constant makes the entire polynomial strictly positive for every real \(x\). Therefore, it can never equal zero.

a) The substitution \(u = x^2\) gives \(u = -1\) or \(u = -2\), neither of which can equal \(x^2\) for real \(x\). Therefore, there are no real zeros. b) All even-power terms are nonnegative for real \(x\), and the positive constant term makes the total strictly positive.

- Set the two function expressions equal at an intersection point. - Move all terms to one side so the equation equals zero. - Look for a common factor before factoring the quadratic part. - Substitute each x-coordinate into either original function to find its y-coordinate.

1. At an intersection point, \(f(x) = g(x)\): \(x^3 + 2x^2 - 4x + 6 = x^2 - 2x + 6\). 2. Move all terms to one side: \(x^3 + x^2 - 2x = 0\). 3. Factor: \(x(x^2 + x - 2) = x(x + 2)(x - 1) = 0\). Thus, the x-coordinates are \(x = -2\), \(x = 0\), and \(x = 1\). 4. Substitute into \(g(x)\): \(g(-2) = 14\), \(g(0) = 6\), and \(g(1) = 5\). 5. The intersection points are \((-2, 14)\), \((0, 6)\), and \((1, 5)\).

\((-2, 14)\), \((0, 6)\), and \((1, 5)\)

- Begin by setting the polynomial equal to zero and factoring. - Factor out the greatest power of \(x\) common to every term. - Read multiplicities from the exponents in complete factored form. - A factor \(x^2\) produces a zero at \(x = 0\) with multiplicity 2.

1. Factor out the greatest common factor: \(p(x) = x^2(x^2 - 2x - 3)\). 2. Factor the quadratic: \(x^2 - 2x - 3 = (x - 3)(x + 1)\). 3. Therefore, \(p(x) = x^2(x - 3)(x + 1)\). 4. The zero \(x = 0\) has multiplicity 2. The zeros \(x = 3\) and \(x = -1\) each have multiplicity 1.

\(x = -1\) with multiplicity 1; \(x = 0\) with multiplicity 2; \(x = 3\) with multiplicity 1

- A product equals zero when at least one factor equals zero. - Analyze the two factors separately. - A positive equation of the form \(x^2 = a\) has two real solutions. - Read multiplicity from the exponent on each linear factor.

1. Use the zero-product property: \(x^4 = 0\) or \(x^2 - 81 = 0\). 2. The equation \(x^4 = 0\) gives \(x = 0\). Because the factor is \(x^4\), this zero has multiplicity 4. 3. Factor the difference of squares: \(x^2 - 81 = (x - 9)(x + 9)\). This gives the zeros \(x = 9\) and \(x = -9\), each with multiplicity 1. 4. Therefore, the function has three distinct real zeros.

Three distinct real zeros: \(x = -9\) with multiplicity 1 \(x = 0\) with multiplicity 4 \(x = 9\) with multiplicity 1

- Solutions of an equation with only even powers occur in opposite pairs. - Use the solutions to write factors. - Pair opposite roots to form expressions of the form \(x^2 - c\). - Expand and compare coefficients.

1. Because the equation contains only even powers of \(x\), the opposite values \(-\sqrt{2}\) and \(-5\) are also solutions. 2. Write the equation in factored form: \((x - \sqrt{2})(x + \sqrt{2})(x - 5)(x + 5) = 0\). 3. Combine conjugate factors: \((x^2 - 2)(x^2 - 25) = 0\). 4. Expand: \(x^4 - 27x^2 + 50 = 0\). 5. Therefore, \(a = -27\) and \(b = 50\).

\(a = -27\), \(b = 50\); the equation is \(x^4 - 27x^2 + 50 = 0\).

- Use each solution to form a factor. - Pair opposite roots and apply the difference-of-squares pattern. - Expand the two quadratic factors.

1. Use the given solutions to write factors: \((x - \sqrt{5})(x + \sqrt{5})(x - \sqrt{2})(x + \sqrt{2}) = 0\). 2. Pair opposite roots: \((x^2 - 5)(x^2 - 2) = 0\). 3. Expand: \(x^4 - 7x^2 + 10 = 0\).

\(x^4 - 7x^2 + 10 = 0\)

- Determine the symmetry of a polynomial with only even powers. - A double zero produces a squared linear factor. - Use the corresponding opposite zero. - Expand the factored form and compare coefficients.

1. The function is even, so a double zero at \(x = 3\) also gives a double zero at \(x = -3\). 2. Write the factored form: \(f(x) = (x - 3)^2(x + 3)^2\). 3. Combine the factors: \(f(x) = (x^2 - 9)^2\). 4. Expand: \(f(x) = x^4 - 18x^2 + 81\). 5. Therefore, \(a = -18\), \(b = 81\), and the real solution set is \(\{-3, 3\}\).

\(a = -18\), \(b = 81\), and the real solution set is \(\{-3, 3\}\).

- Separate the parameter-dependent and parameter-independent parts. - Factor the function before counting distinct zeros. - Analyze when an equation of the form \(x^2=c\) has real solutions. - A horizontal shift replaces \(x\) with \(x-h\).

1. Rewrite \(f_a(x)=a(x^4-x^2)-x^2\). A point is independent of \(a\) when \(x^4-x^2=0\), so \(x=0\) or \(x=\pm1\). The corresponding points are \((0, 0)\), \((1, -1)\), and \((-1, -1)\). 2. Factor \(f_a(x)=x^2(ax^2-a-1)\). Besides \(x=0\), additional real zeros exist when \(x^2=\frac{a+1}{a}\) is positive. Exactly one distinct real zero occurs when \(\frac{a+1}{a}\le0\), which gives \(-1\le a<0\). 3. Only even powers of \(x\) occur, so \(f_a(-x)=f_a(x)\). Shifting the graph \(2\) units right gives \(g_a(x)=f_a(x-2)=a(x-2)^4-(a+1)(x-2)^2\), whose axis of symmetry is \(x=2\).

a) \((0, 0)\), \((1, -1)\), and \((-1, -1)\) b) \(-1\le a<0\) c) \(g_a(x)=a(x-2)^4-(a+1)(x-2)^2\)

- Origin symmetry implies that the polynomial is an odd function. - Zeros of an odd function occur in opposite pairs, along with \(x=0\). - Use factored form to build a polynomial from its zeros. - Include a nonzero leading coefficient to describe all possible functions.

1. Origin symmetry means the polynomial is odd. Therefore, if \(1\) and \(2\) are zeros, then \(-1\), \(-2\), and \(0\) are also zeros. 2. A fifth-degree polynomial with these five zeros has the form \(f_a(x)=a x(x-1)(x+1)(x-2)(x+2)\), where \(a\ne0\). 3. Simplify: \(f_a(x)=a x(x^2-1)(x^2-4)\) \(=a(x^5-5x^3+4x)\). Every function in this family has degree \(5\), origin symmetry, and the required x-intercepts.

\(f_a(x)=a(x^5-5x^3+4x)\), where \(a\in\mathbb R\setminus\{0\}\)

- Factor each quadratic expression completely. - How does odd or even multiplicity affect a graph at a zero? - Distinguish between crossing the x-axis and touching it without crossing.

1. Factor the difference of squares: \(x^2 - 1 = (x - 1)(x + 1)\). 2. Factor the perfect-square trinomial: \(x^2 - 10x + 25 = (x - 5)^2\). 3. The complete factored form is \(h(x) = (x - 1)(x + 1)(x - 5)^2\). 4. The zeros are \(x = -1\), \(x = 1\), and \(x = 5\). The zeros \(x = -1\) and \(x = 1\) each have multiplicity 1, while \(x = 5\) has multiplicity 2. 5. A graph crosses the x-axis at a zero with odd multiplicity and touches without crossing at a zero with even multiplicity. Therefore, the graph crosses at \(x = -1\) and \(x = 1\), and it touches at \(x = 5\).

The graph crosses the x-axis at \(x = -1\) and \(x = 1\), where the zeros have multiplicity 1. It touches the x-axis at \(x = 5\), where the zero has multiplicity 2.

- Distinguish between the greatest possible number of zeros and the actual number. - Try to construct a cubic whose remaining quadratic factor has no real zeros. - When a power is squared, multiply its exponent by \(2\).

1. When \(n = 8\), the function is \(p(x) = x^8 - x + 10\). Its degree is \(8\), so it can have at most \(8\) real zeros. 2. The claim in b) is false because the degree gives a maximum, not the exact number of real zeros. For example, \(q(x) = x^3 + x = x(x^2 + 1)\). Since \(x^2 + 1\) has no real zeros, \(q\) has only the real zero \(x = 0\). 3. For c), \((3x^3 - 2)^2 = 9x^6 - 12x^3 + 4\). The degree is \(6\), so the function can have at most \(6\) real zeros.

a) At most \(8\) real zeros b) The claim is false. For example, \(q(x) = x^3 + x\) has only one real zero, \(x = 0\). c) Degree \(6\); at most \(6\) real zeros

- Expand each product and combine like terms. - The degree of the simplified polynomial determines whether it is quadratic. - Factor the quadratic to find its zeros. - A zero of odd multiplicity produces a sign change.

1. Expand the first product: \((x+1)(2x-4)=2x^2-2x-4\). 2. Expand the square: \((x-2)^2=x^2-4x+4\). 3. Subtract and combine like terms: \(f(x)=2x^2-2x-4-(x^2-4x+4)=x^2+2x-8\). The leading term is \(x^2\), so \(f\) is quadratic. 4. Factor: \(x^2+2x-8=(x+4)(x-2)\). The zeros are \(x=-4\) and \(x=2\), each with multiplicity \(1\). 5. A polynomial changes sign at each zero of odd multiplicity, so \(f\) changes sign at \(x=-4\) and \(x=2\).

\(f(x)=x^2+2x-8\). The function changes sign at \(x=-4\) and \(x=2\).

- Set the function value equal to zero. - The exponents \(6\) and \(3\) suggest a substitution involving \(x^3\). - Solve the resulting quadratic equation first. - Substitute back and solve each cubic equation over the real numbers.

1. Set the function equal to zero: \(x^6 + 26x^3 - 27 = 0\). 2. Substitute \(u = x^3\). Then \(x^6 = u^2\), so the equation becomes \(u^2 + 26u - 27 = 0\). 3. Factor the quadratic: \(u^2 + 26u - 27 = (u - 1)(u + 27)\). Thus, \(u = 1\) or \(u = -27\). 4. Substitute back. From \(x^3 = 1\), \(x = 1\). From \(x^3 = -27\), \(x = -3\). 5. Therefore, the real zeros are \(x = -3\) and \(x = 1\).

The real zeros are \(x = -3\) and \(x = 1\).

- Look for a greatest common factor first. - Use difference-of-squares and perfect-square identities where applicable. - A repeated linear factor produces a zero with the same multiplicity as its exponent. - Odd multiplicity gives a sign change; even multiplicity does not.

1. For a), factor out \(x\) and then use the difference of squares: \(f(x) = x(x^2 - 4) = x(x - 2)(x + 2)\). The zeros \(x = -2\), \(x = 0\), and \(x = 2\) each have multiplicity 1, so the function changes sign at each zero. 2. For b), factor out \(x^2\): \(g(x) = x^2(x^2 - 2x + 1) = x^2(x - 1)^2\). The zeros \(x = 0\) and \(x = 1\) each have multiplicity 2, so the function does not change sign at either zero. 3. For c), factor the difference of squares: \(h(x) = (x - 3)(x + 3)(x + 3) = (x - 3)(x + 3)^2\). The zero \(x = 3\) has multiplicity 1 and a sign change. The zero \(x = -3\) has multiplicity 2 and no sign change.

a) \(f(x) = x(x - 2)(x + 2)\); \(x = -2\), \(0\), and \(2\) each have multiplicity 1 and a sign change b) \(g(x) = x^2(x - 1)^2\); \(x = 0\) and \(x = 1\) each have multiplicity 2 and no sign change c) \(h(x) = (x - 3)(x + 3)^2\); \(x = 3\) has multiplicity 1 and a sign change; \(x = -3\) has multiplicity 2 and no sign change

- A zero \(x = r\) corresponds to a factor \(x - r\). - Use linear-factor form to build the polynomial. - The sum of the multiplicities equals the degree. - If there are fewer distinct zeros than the degree, at least one zero must be repeated.

1. Build each function from linear factors. The sum of the multiplicities must equal the degree. 2. For a), three distinct zeros in a degree-\(3\) polynomial can each have multiplicity 1. One example is \(f(x) = x(x + 4)(x - 1)\). 3. For b), the two given zeros must have multiplicities that add to \(4\). One example is \(f(x) = (x + 2)^2(x - 5)^2\). 4. For c), a degree-\(3\) polynomial with exactly two distinct zeros must have one simple zero and one double zero. Choosing \(x = 3\) as the double zero and \(x = 0\) as the other zero gives \(f(x) = x(x - 3)^2\).

a) \(f(x) = x(x + 4)(x - 1)\) b) \(f(x) = (x + 2)^2(x - 5)^2\) c) \(f(x) = x(x - 3)^2\)

- Factor the polynomial completely. - A shared point must be independent of \(k\). - Tangency at the x-axis corresponds to a zero of even multiplicity. - For origin symmetry, test whether \(f(-x)=-f(x)\).

1. Factor the family: \(f_k(x)=(x^2-4)(x-k)=(x-2)(x+2)(x-k)\). 2. The factors \(x-2\) and \(x+2\) are present for every \(k\), so every graph contains \((2, 0)\) and \((-2, 0)\). 3. The graph is tangent to the x-axis when one of these zeros is repeated. If \(k=2\), then \(x=2\) has multiplicity \(2\), giving tangency at \((2, 0)\). If \(k=-2\), then \(x=-2\) has multiplicity \(2\), giving tangency at \((-2, 0)\). 4. When \(k=0\), \(f_0(x)=x^3-4x\). Then \(f_0(-x)=-x^3+4x=-f_0(x)\), so \(f_0\) is odd and its graph has rotational symmetry about the origin.

a) \((-2, 0)\) and \((2, 0)\) b) \(k=-2\) at \((-2, 0)\); \(k=2\) at \((2, 0)\) c) \(f_0(-x)=-f_0(x)\), so the graph is symmetric about the origin.

- Factor the polynomial completely. - What does an even-multiplicity zero tell you about the graph at the x-axis? - Compare \(f_a(-x)\) with \(-f_{-a}(x)\). - Recall the algebraic condition for a \(180^\circ\) rotation about the origin.

1. Factor the polynomial: \(f_a(x)=(x-2a)^2(x+a)\). 2. Therefore, \(x=2a\) is a zero of even multiplicity for every nonzero \(a\). A graph touches, rather than crosses, the x-axis at a zero of even multiplicity. Thus, every graph is tangent to the x-axis at \((2a, 0)\). 3. To compare the two graphs, evaluate \(f_a(-x)\): \(f_a(-x)=-x^3-3ax^2+4a^3\). 4. Also, \(-f_{-a}(x)=-\left(x^3+3ax^2-4a^3\right)=-x^3-3ax^2+4a^3\). 5. Since \(f_a(-x)=-f_{-a}(x)\) for every \(x\), a \(180^\circ\) rotation about the origin maps the graph of \(f_a\) onto the graph of \(f_{-a}\).

a) Since \(f_a(x)=(x-2a)^2(x+a)\), every graph is tangent to the x-axis at \((2a, 0)\). b) \(f_a(-x)=-f_{-a}(x)\), so the two graphs have rotational symmetry about the origin with respect to each other.

- For a shared point, the term multiplied by \(k\) must not affect the output. - Factor \(f_k(x)\) before analyzing its zeros. - One zero is already fixed at \(x=0\). - Use the discriminant to determine when the quadratic factor has a repeated zero, then check each resulting value of \(k\).

1. A point shared by every graph must have a function value independent of \(k\). The parameter term is zero when \(x^2-6x=x(x-6)=0\), so \(x=0\) or \(x=6\). 2. Evaluate the corresponding y-values: \(f_k(0)=0\) and \(f_k(6)=216\). Therefore, the shared points are \((0, 0)\) and \((6, 216)\). 3. Factor the equation for the x-intercepts: \(f_k(x)=x(x^2+kx-6k)\). One zero is always \(x=0\). 4. For exactly two distinct x-intercepts, the quadratic factor must have one repeated nonzero zero. Its discriminant must equal zero: \(k^2+24k=0\), so \(k=0\) or \(k=-24\). 5. When \(k=0\), \(f_0(x)=x^3\), which has only one distinct zero. When \(k=-24\), \(f_{-24}(x)=x(x-12)^2\), which has the two distinct zeros \(0\) and \(12\). Therefore, \(k=-24\).

1. \((0, 0)\) and \((6, 216)\) 2. \(k=-24\)

- Identify where the graph crosses or touches the x-axis. - Crossing indicates odd multiplicity; touching without crossing indicates even multiplicity. - Match each zero with its linear factor and exponent. - Use the end behavior to determine the sign of the leading coefficient.

1. The graph has zeros at \(x = -1\), \(x = 1\), and \(x = 3\). 2. At \(x = -1\), the graph crosses the x-axis, so the zero has odd multiplicity. At \(x = 1\), the graph touches the x-axis without crossing, so the zero has even multiplicity. At \(x = 3\), the graph crosses again, so that zero has odd multiplicity. 3. These behaviors give the form \(f(x) = a(x - 1)^2(x + 1)(x - 3)\). 4. Both ends of the graph rise, so the leading coefficient is positive. Among the choices, this gives \(a = 1\). 5. Therefore, choice **A** matches the graph.

**A**: \(f(x) = (x - 1)^2(x + 1)(x - 3)\)

- Find the zeros directly from each factored expression. - Identify which zeros are simple and which are repeated. - Odd multiplicity means the graph crosses the x-axis. - Even multiplicity means the graph touches without crossing. - Compare the number and type of x-intercepts in each graph.

1. For \(f\), the zero \(x = -1\) has multiplicity 2, so the graph touches the x-axis there. The zero \(x = 2\) has multiplicity 1, so the graph crosses there. This matches Graph B. 2. For \(g\), the zero \(x = -1\) has multiplicity 1, so the graph crosses there. The zero \(x = 2\) has multiplicity 2, so the graph touches there. This matches Graph C. 3. For \(h\), the zeros \(x = -1\), \(x = 1\), and \(x = 2\) each have multiplicity 1, so the graph crosses the x-axis at all three zeros. This matches Graph A.

\(f\) matches Graph B. \(g\) matches Graph C. \(h\) matches Graph A.

- Identify where the graph crosses or touches the x-axis. - Crossing indicates a simple zero; touching indicates an even-multiplicity zero. - Write a factored expression with an unknown leading coefficient. - Use the exact y-intercept to solve for the leading coefficient.

1. The graph crosses the x-axis at \(x = -1\), so that zero has multiplicity 1. The graph touches the x-axis at \(x = 2\), so that zero has multiplicity 2. 2. A cubic with these zeros has the form \(f(x) = a(x + 1)(x - 2)^2\). 3. The graph has y-intercept \((0, -2)\). Substitute this point: \(-2 = a(0 + 1)(0 - 2)^2 = 4a\). 4. Therefore, \(a = -0.5\), and \(f(x) = -0.5(x + 1)(x - 2)^2\).

a) \(x = -1\) with multiplicity 1; \(x = 2\) with multiplicity 2 b) \(f(x) = -0.5(x + 1)(x - 2)^2\)

- Locate where the graph crosses and where it touches the x-axis. - A simple zero corresponds to the unsquared factor; a double zero corresponds to the squared factor. - Read the exact y-intercept from the graph. - Substitute the y-intercept into the factored form and solve for \(c\).

1. The graph crosses the x-axis at \(x = -1\), so this is the simple zero and \(a = -1\). The graph touches the x-axis at \(x = 2\), so this is the double zero and \(b = 2\). 2. The function therefore has the form \(f(x) = c(x + 1)(x - 2)^2\). 3. The y-intercept is \((0, 2)\). Substitute this point: \(2 = c(0 + 1)(0 - 2)^2 = 4c\). 4. Thus, \(c = 0.5\), and the complete function is \(f(x) = 0.5(x + 1)(x - 2)^2\).

1. \(a = -1\) and \(b = 2\) 2. \(c = 0.5\)

- Identify where each graph crosses or touches the x-axis. - Crossing indicates odd multiplicity; touching indicates even multiplicity. - Match the zeros and their multiplicities to the factors. - Use the right-end behavior to determine the sign of the leading coefficient.

1. Graph A crosses at \(x = -1\) and touches at \(x = 2\). Its right end rises, so the leading coefficient is positive. This matches equation 1, \(p(x) = \frac{1}{4}(x + 1)(x - 2)^2\). 2. Graph B touches at \(x = -1\) and crosses at \(x = 2\). Its right end falls, so the leading coefficient is negative. This matches equation 2, \(q(x) = -\frac{1}{4}(x + 1)^2(x - 2)\). 3. Graph C crosses at \(x = -1\), \(x = 0\), and \(x = 2\). Its right end rises, so the leading coefficient is positive. This matches equation 3, \(r(x) = \frac{1}{4}x(x + 1)(x - 2)\). 4. Equation 4 has the same zeros as equation 3 but the opposite end behavior, so it does not match any displayed graph.

Graph A: equation 1 Graph B: equation 2 Graph C: equation 3

- Identify where each graph crosses and where it touches the x-axis. - A simple zero gives a crossing; a double zero gives a touch. - Use the sign of the leading coefficient to determine the right-end behavior. - Combine multiplicity and end behavior to make each match.

1. Functions (1) and (2) have a simple zero at \(x = -2\) and a double zero at \(x = 1\). Functions (3) and (4) have a double zero at \(x = -2\) and a simple zero at \(x = 1\). 2. Graph A crosses at \(x = -2\), touches at \(x = 1\), and rises to the right. It matches function (1). 3. Graph B crosses at \(x = -2\), touches at \(x = 1\), and falls to the right. It matches function (2). 4. Graph C touches at \(x = -2\), crosses at \(x = 1\), and falls to the right. It matches function (4). 5. Function (3) is unused because it would touch at \(x = -2\), cross at \(x = 1\), and rise to the right.

Graph A: function (1) Graph B: function (2) Graph C: function (4) Function (3) is not used.

- Identify the double zero, simple zero, and triple zero shown by the graphs. - A zero where the graph only touches the x-axis corresponds to an even multiplicity. - A flattened crossing at an x-intercept indicates multiplicity 3 in this problem. - Read the zeros first, then use one additional exact point on each graph to find the scale factor.

1. Graph a touches the x-axis at \(x = 1\), indicating a double zero, and crosses at \(x = -2\), indicating a simple zero. Therefore, graph a matches \(p(x) = a(x - b)^2(x - c)\), with \(b = 1\) and \(c = -2\). 2. Use the point \((-1, -1)\) on graph a: \(-1 = a(-1 - 1)^2(-1 + 2) = 4a\). Thus, \(a = -0.25\). 3. Graph b crosses the x-axis with flattening at \(x = 2\), indicating a triple zero. Thus, it matches \(q(x) = d(x - e)^3 + f\), with \(e = 2\) and \(f = 0\). 4. Use the point \((3, 1)\) on graph b: \(1 = d(3 - 2)^3\), so \(d = 1\).

a) Graph a represents \(p\), and graph b represents \(q\). b) \(a = -0.25\), \(b = 1\), \(c = -2\), \(d = 1\), \(e = 2\), \(f = 0\)

- Determine where the graph crosses and where it touches the x-axis. - Crossing indicates odd multiplicity; touching indicates even multiplicity. - Write factored form with an unknown leading coefficient. - Substitute the coordinates of \(P\) to solve for the coefficient.

1. The graph crosses the x-axis at \(x = -3\), so that zero has multiplicity 1. It touches the x-axis at \(x = 1\), so that zero has multiplicity 2. 2. A cubic with these zeros has the form \(f(x) = a(x + 3)(x - 1)^2\). 3. Substitute \(P(2, 1)\): \(1 = a(2 + 3)(2 - 1)^2 = 5a\). 4. Therefore, \(a = \frac{1}{5} = 0.2\), and \(f(x) = \frac{1}{5}(x + 3)(x - 1)^2\).

1. \(x = -3\) with multiplicity 1; \(x = 1\) with multiplicity 2 2. \(f(x) = \frac{1}{5}(x + 3)(x - 1)^2\)

- Identify whether each zero is simple, double, or triple from the graph behavior. - Convert each zero and multiplicity into a linear factor and exponent. - Use the sum of multiplicities to obtain the least possible degree. - Substitute an exact grid point to determine the scale factor.

1. For graph a), the graph crosses at \(x = -1\), so that zero has multiplicity 1. It touches at \(x = 2\), so that zero has multiplicity 2. The least possible degree is \(3\), and \(f(x) = a(x + 1)(x - 2)^2\). 2. The y-intercept is \((0, 1)\). Substituting gives \(1 = a(1)(4)\), so \(a = \frac{1}{4}\). Thus, \(f(x) = \frac{1}{4}(x + 1)(x - 2)^2\). 3. For graph b), the graph crosses the x-axis with a horizontal tangent at \(x = -1\), so that zero has multiplicity 3. It crosses normally at \(x = 2\), so that zero has multiplicity 1. The least possible degree is \(4\), and \(g(x) = a(x + 1)^3(x - 2)\). 4. The y-intercept is \((0, -1)\). Substituting gives \(-1 = a(1)^3(-2)\), so \(a = \frac{1}{2}\). Thus, \(g(x) = \frac{1}{2}(x + 1)^3(x - 2)\).

a) \(f(x) = \frac{1}{4}(x + 1)(x - 2)^2\) b) \(g(x) = \frac{1}{2}(x + 1)^3(x - 2)\)

- Determine whether the graph crosses or touches at each x-intercept. - Translate each zero and multiplicity into a linear factor and exponent. - Substitute the marked point to determine the leading coefficient. - Expand the factored form using the distributive property.

1. At \(x = -1\), the graph touches the x-axis without crossing, so \(x = -1\) is a double zero. At \(x = 2\), the graph crosses the x-axis, so \(x = 2\) is a simple zero. 2. The factored form has the shape \(f(x) = a(x + 1)^2(x - 2)\). 3. Substitute \(P(0, -2)\): \(-2 = a(0 + 1)^2(0 - 2) = -2a\), so \(a = 1\). 4. Therefore, \(f(x) = (x + 1)^2(x - 2)\). 5. Expand: \((x^2 + 2x + 1)(x - 2) = x^3 - 3x - 2\).

1. \(x = -1\) is a double zero; \(x = 2\) is a simple zero. 2. Factored form: \(f(x) = (x + 1)^2(x - 2)\) Standard form: \(f(x) = x^3 - 3x - 2\)

- Read the exact x-intercepts and y-intercept from the graph. - Touching indicates an even-multiplicity zero; crossing indicates an odd-multiplicity zero. - Write the factored form with an unknown leading coefficient. - Substitute the y-intercept to determine that coefficient.

1. The graph touches the x-axis at \(x = -2\), so \(x = -2\) is a double zero. It crosses the x-axis at \(x = 1\), so \(x = 1\) is a simple zero. 2. The factored form is \(f(x) = a(x + 2)^2(x - 1)\). 3. The y-intercept is \((0, -2)\). Substitute it: \(-2 = a(0 + 2)^2(0 - 1) = -4a\). 4. Therefore, \(a = 0.5\), so \(f(x) = 0.5(x + 2)^2(x - 1)\). 5. Expanding gives \(f(x) = 0.5x^3 + 1.5x^2 - 2\).

\(f(x) = 0.5(x + 2)^2(x - 1)\), or \(f(x) = 0.5x^3 + 1.5x^2 - 2\) in standard form.

- Use symmetry to identify whether only even or only odd powers should appear. - Read the zeros from each graph and write a factored form. - Crossing x-intercepts correspond to simple factors. - Substitute the given point to determine the scale factor. - Expand the factored expression into the requested standard form.

1. The zeros of \(f\) are \(x = -2\), \(x = -1\), \(x = 1\), and \(x = 2\). Y-axis symmetry pairs the zeros, so \(f(x) = A(x^2 - 1)(x^2 - 4)\). 2. Substitute \(P(0, 2)\): \(2 = A(-1)(-4) = 4A\), so \(A = 0.5\). Expanding gives \(f(x) = 0.5x^4 - 2.5x^2 + 2\). 3. The zeros of \(g\) are \(x = -3\), \(x = 0\), and \(x = 3\). Origin symmetry is consistent with \(g(x) = Bx(x^2 - 9)\). 4. Substitute \(Q(2, -2)\): \(-2 = B \cdot 2(4 - 9) = -10B\), so \(B = 0.2\). Expanding gives \(g(x) = 0.2x^3 - 1.8x\).

\(f(x) = 0.5x^4 - 2.5x^2 + 2\) \(g(x) = 0.2x^3 - 1.8x\)

- First identify which graph falls to the right. - A simple zero produces a crossing; a double zero produces a touch. - Determine the location and multiplicity of each zero from the factors. - Use the y-intercept as an optional check.

1. Function \(g\) has a negative leading coefficient, so its graph rises to the left and falls to the right. It touches at \(x = -2\) and crosses at \(x = 1\). This matches Graph A. 2. Function \(f\) has a positive leading coefficient, crosses at \(x = -2\), and touches at \(x = 1\). This matches Graph B. 3. Function \(h\) also has a positive leading coefficient, but it touches at \(x = -2\) and crosses at \(x = 1\). This matches Graph C.

Graph A: \(g(x) = -(x - 1)(x + 2)^2\) Graph B: \(f(x) = (x - 1)^2(x + 2)\) Graph C: \(h(x) = (x - 1)(x + 2)^2\)

- Compare how a graph behaves at a zero of multiplicity 1 with how it behaves at a zero of multiplicity 2. - Use the factored form \(f(x) = a(x - x_1)^{m_1}(x - x_2)^{m_2}\). - Read the y-intercept from the graph. - Substitute the y-intercept into your factored form to determine \(a\).

1. The graph touches the x-axis at \(x = -1\), so \(x = -1\) is a zero of multiplicity 2. The graph crosses the x-axis at \(x = 3\), so \(x = 3\) is a zero of multiplicity 1. 2. A factored form with these zeros is \(f(x) = a(x + 1)^2(x - 3)\). 3. The y-intercept is \((0, 1.5)\). Substitute this point: \(1.5 = a(0 + 1)^2(0 - 3) = -3a\), so \(a = -0.5\). 4. Therefore, \(f(x) = -0.5(x + 1)^2(x - 3)\).

a) The zero \(x = -1\) has multiplicity 2, and the zero \(x = 3\) has multiplicity 1. b) \(f(x) = -0.5(x + 1)^2(x - 3)\)

- Use factored form when all zeros are known. - Include an unknown nonzero leading factor. - Substitute the additional point to find that factor. - Expand the factors carefully. - Check that the expanded form retains all three zeros.

1. The zeros give \(f(x)=a(x+2)(x-1)(x-3)\). 2. Using \(P(-1, 4)\), \(4=a(1)(-2)(-4)=8a\), so \(a=\frac{1}{2}\). 3. Thus, the factored form is \(f(x)=\frac{1}{2}(x+2)(x-1)(x-3)\). 4. Expanding gives \(f(x)=\frac{1}{2}x^3-x^2-\frac{5}{2}x+3\).

Factored form: \(f(x)=\frac{1}{2}(x+2)(x-1)(x-3)\) Expanded form: \(f(x)=\frac{1}{2}x^3-x^2-\frac{5}{2}x+3\)

- Identify each zero and decide whether the graph crosses or touches the x-axis there. - Use the left- and right-end behavior to determine the sign of the leading coefficient. - Compare the zeros in the graph with the values in the factors. - Substitute \(x = 0\) as an additional check.

1. The graph touches the x-axis at \(x = -2\) and \(x = 2\), so both zeros have even multiplicity. This requires the factors \((x + 2)^2\) and \((x - 2)^2\). 2. Both ends of the graph fall, so the polynomial has even degree and a negative leading coefficient. 3. Only choice c) has double zeros at \(x = -2\) and \(x = 2\) together with a negative leading coefficient. 4. As a check, \(p(0) = -0.1 \cdot 2^2 \cdot (-2)^2 = -1.6\), which agrees with the graph. 5. Therefore, choice c) is correct.

c) \(p(x) = -0.1(x + 2)^2(x - 2)^2\)

- Use a visible point and the origin to find the line’s slope. - Set the two function expressions equal at an intersection. - Move all terms to one side and factor.

1. The line passes through \((0,0)\) and \((1,3)\), so its slope is \(3\). Therefore, \(g(x)=3x\). 2. At an intersection, \(x^3=3x\). 3. The graph suggests solutions near \(-1.7\), \(0\), and \(1.7\). 4. Algebraically, \(x^3-3x=x(x^2-3)=0\). Thus, \(x=0\) or \(x=\pm\sqrt{3}\), giving \(x\approx -1.732, 0.000, 1.732\).

1. \(g(x)=3x\) 2. \(x^3=3x\) 3. \(x\approx -1.732, 0.000, 1.732\)

- Substitute \(x = 0\) to find each y-intercept. - Factor when possible to identify the zeros. - A graph touches the x-axis at a zero of even multiplicity. - Use y-axis symmetry to identify the functions containing only even powers of \(x\).

1. Factor \(p(x)\): \(p(x) = (x + 1)(x - 1)(x - 2)\). It has three simple zeros at \(x = -1\), \(x = 1\), and \(x = 2\), and its y-intercept is \(2\). This matches Graph B. 2. Function \(q\) is even, so its graph is symmetric about the y-axis. Its y-intercept is \(1\), and its negative leading coefficient makes both ends fall. This matches Graph A. 3. Factor \(r(x)\): \(r(x) = 0.5(x - 1)^2(x + 2)\). It has a double zero at \(x = 1\), a simple zero at \(x = -2\), and a y-intercept of \(1\). This matches Graph D. 4. Factor \(s(x)\): \(s(x) = (x - 2)(x - 1)(x + 1)(x + 2)\). It has simple zeros at \(x = -2\), \(x = -1\), \(x = 1\), and \(x = 2\), and its y-intercept is \(4\). This matches Graph C.

1. Graph B 2. Graph A 3. Graph D 4. Graph C

- A zero of multiplicity 1 produces a crossing of the x-axis. - A zero of multiplicity 2 produces a touch at the x-axis. - Use the degree and the sign of the leading coefficient to determine end behavior. - Factor each equation before comparing it with the graphs.

1. Factor \(p(x)\): \(p(x) = 0.5x(x - 2)(x + 2)\). It has three simple zeros at \(x = -2\), \(x = 0\), and \(x = 2\). Its left end falls and its right end rises. This matches Graph C. 2. Factor \(q(x)\): \(q(x) = -0.25x^2(x - 2)(x + 2)\). It has a double zero at \(x = 0\) and simple zeros at \(x = -2\) and \(x = 2\). Because its degree is even and its leading coefficient is negative, both ends fall. This matches Graph A. 3. Function \(r(x) = (x - 1)^2(x + 2)\) has a double zero at \(x = 1\) and a simple zero at \(x = -2\). This matches Graph B.

1. Graph C 2. Graph A 3. Graph B

- Look for a substitution that turns the equation into a quadratic equation. - After solving for the substituted variable, replace it with the original expression. - Each equation of the form \(x^3 = a\) has exactly one real solution. - Use the x-intercepts of the graph to check your results.

1. Because the exponents are \(6\) and \(3\), let \(z = x^3\). Then \(x^6 = z^2\). 2. The equation \(h(x) = 0\) becomes \(z^2 - 7z - 8 = 0\). 3. Factor: \(z^2 - 7z - 8 = (z - 8)(z + 1)\). Thus, \(z = 8\) or \(z = -1\). 4. Substitute back. From \(x^3 = 8\), \(x = 2\). From \(x^3 = -1\), \(x = -1\). 5. The graph crosses the x-axis at \(x = -1\) and \(x = 2\), confirming both zeros.

The real zeros are \(x = -1\) and \(x = 2\).

- Set the two function expressions equal to find the x-coordinates. - Look for a common factor in the resulting equation. - Substitute each x-coordinate into the simpler function to find its y-coordinate. - Use the graph to check whether the calculated points are reasonable.

1. Set the functions equal: \(x^3 - 3x + 1 = x + 1\). 2. Rewrite and factor: \(x^3 - 4x = 0\), so \(x(x^2 - 4) = x(x - 2)(x + 2) = 0\). 3. The x-coordinates are \(x = -2\), \(x = 0\), and \(x = 2\). 4. Substitute into \(g(x) = x + 1\): \(g(-2) = -1\), \(g(0) = 1\), and \(g(2) = 3\). 5. The intersection points are \((-2, -1)\), \((0, 1)\), and \((2, 3)\). 6. The graph shows the line and cubic intersecting at these three points.

a) \((-2, -1)\), \((0, 1)\), and \((2, 3)\) b) The graph confirms all three intersection points.

- Decide whether the graph crosses or touches the x-axis at each zero. - A touch indicates an even multiplicity, while a crossing indicates an odd multiplicity. - Write a factored equation with an unknown leading factor \(a\). - Substitute the additional marked point to determine \(a\).

1. The graph touches the x-axis at \(x = -3\), so \(x = -3\) is a zero of multiplicity 2. It crosses at \(x = 1\), so \(x = 1\) is a zero of multiplicity 1. 2. The lowest-degree equation has the form \(f(x) = a(x + 3)^2(x - 1)\). 3. Use the marked point \((0, -1.8)\): \(-1.8 = a(0 + 3)^2(0 - 1) = -9a\). 4. Solving gives \(a = 0.2\). 5. Therefore, \(f(x) = 0.2(x + 3)^2(x - 1)\).

\(f(x) = 0.2(x + 3)^2(x - 1)\)

- Use the crossing and touching behavior to determine the multiplicity of each zero. - Write a cubic equation in factored form with an unknown leading factor. - Substitute the point \((1, -2)\) to determine the leading factor.

1. The graph touches the x-axis at \(x = 0\), so \(x = 0\) is a zero of multiplicity 2. It crosses at \(x = 3\), so \(x = 3\) is a zero of multiplicity 1. 2. A cubic equation with these zeros is \(f(x) = ax^2(x - 3)\). 3. Use the point \((1, -2)\): \(-2 = a(1)^2(1 - 3) = -2a\). 4. Thus, \(a = 1\). 5. Therefore, \(f(x) = x^2(x - 3)\), or equivalently \(f(x) = x^3 - 3x^2\).

\(f(x) = x^2(x - 3)\)

- Use each x-intercept to write a linear factor. - Crossing the x-axis indicates a zero of odd multiplicity; use the lowest possible multiplicity here. - Include an unknown leading factor \(a\), then substitute the given point.

1. Because the cubic has three simple zeros, write \(f(x) = a(x + 2)x(x - 2)\). 2. Substitute \(P(1, -3)\): \(-3 = a(1 + 2)(1)(1 - 2) = -3a\). 3. Therefore, \(a = 1\). 4. The equation is \(f(x) = x(x + 2)(x - 2)\), or equivalently \(f(x) = x^3 - 4x\).

\(f(x) = x(x + 2)(x - 2)\)

- Use each x-intercept to write a linear factor. - Substitute \(x = 0\) and use the given y-intercept to determine the leading factor.

1. Three crossings require three simple zeros, so the lowest possible degree is 3. Write \(f(x) = a(x + 1)(x - 1)(x - 2)\). 2. The y-intercept is \((0, 2)\). Substitute this point: \(2 = a(0 + 1)(0 - 1)(0 - 2) = 2a\). 3. Thus, \(a = 1\). 4. Therefore, \(f(x) = (x + 1)(x - 1)(x - 2)\).

\(f(x) = (x + 1)(x - 1)(x - 2)\)

- A graph that touches the x-axis has a zero of even multiplicity. - Use the symmetry of the two zeros to simplify the factored form.

1. Touching the x-axis indicates an even multiplicity. For the lowest possible degree, use multiplicity 2 at each zero. 2. Write \(f(x) = a(x + 2)^2(x - 2)^2 = a(x^2 - 4)^2\). 3. Substitute \(S(0, -4)\): \(-4 = a(0^2 - 4)^2 = 16a\). 4. Thus, \(a = -0.25\). 5. Therefore, \(f(x) = -0.25(x^2 - 4)^2\).

\(f(x) = -0.25(x^2 - 4)^2\)

- Translate each touching point into a zero of even multiplicity. - Use the factored form for two double zeros, then substitute the given point.

1. Each touch indicates a zero of even multiplicity. Because the polynomial has degree 4, both zeros must have multiplicity 2. 2. Write \(f(x) = ax^2(x - 3)^2\). 3. Substitute \(P(1, 4)\): \(4 = a(1)^2(1 - 3)^2 = 4a\). 4. Therefore, \(a = 1\). 5. The equation is \(f(x) = x^2(x - 3)^2\).

\(f(x) = x^2(x - 3)^2\)

- Add the smallest possible multiplicities to determine the lowest possible degree. - Use crossing or touching behavior to choose the exponent of each factor.

1. The touch at \(x = 0\) gives a zero of multiplicity 2. The crossings at \(x = -2\) and \(x = 2\) give simple zeros. The lowest possible degree is \(2 + 1 + 1 = 4\). 2. Write \(f(x) = ax^2(x + 2)(x - 2) = ax^2(x^2 - 4)\). 3. Substitute \(P(1, -3)\): \(-3 = a(1)^2(1^2 - 4) = -3a\). 4. Thus, \(a = 1\). 5. Therefore, \(f(x) = x^2(x^2 - 4)\), or equivalently \(f(x) = x^4 - 4x^2\).

\(f(x) = x^2(x^2 - 4)\)

- Identify every zero and decide whether it is simple or double. - Use the y-intercept to determine the leading factor, then check its sign against the end behavior.

1. The graph crosses the x-axis at \(x = -1\) and \(x = 2\), so these are simple zeros. It touches at \(x = 1\), so \(x = 1\) is a double zero. 2. Write \(f(x) = a(x + 1)(x - 2)(x - 1)^2\). 3. The y-intercept is \((0, 2)\). Substitute it: \(2 = a(1)(-2)(-1)^2 = -2a\). 4. Therefore, \(a = -1\). 5. The equation is \(f(x) = -(x + 1)(x - 2)(x - 1)^2\).

\(f(x) = -(x + 1)(x - 2)(x - 1)^2\)

- Match each form to the number and multiplicities of the zeros shown by the graph. - A graph touches at a zero of even multiplicity and crosses at a zero of odd multiplicity. - Use a readable point, such as the y-intercept, to determine the leading factor.

1. Graph \(f\) has one touching zero and one crossing zero, so it matches \(p\). Graph \(g\) has two touching zeros, so it matches \(q\). 2. For graph \(f\), the double zero is \(x_1 = -1\) and the simple zero is \(x_2 = 2\). Thus, \(p(x) = a(x + 1)^2(x - 2)\). 3. The y-intercept of graph \(f\) is \((0, -2)\). Substitution gives \(-2 = a(1)^2(-2)\), so \(a = 1\). 4. For graph \(g\), the double zeros are \(x_3 = -2\) and \(x_4 = 2\). Thus, \(q(x) = b(x + 2)^2(x - 2)^2\). 5. The y-intercept of graph \(g\) is \((0, 2)\). Substitution gives \(2 = b(2)^2(-2)^2 = 16b\), so \(b = \frac{1}{8} = 0.125\).

Graph \(f\) matches \(p\), with \(a = 1\), \(x_1 = -1\), and \(x_2 = 2\). Graph \(g\) matches \(q\), with \(b = \frac{1}{8}\), \(x_3 = -2\), and \(x_4 = 2\).

- Relate the exponent of \((x - 1)\) to the zero's multiplicity and graph behavior. - Use the degree and leading coefficient to determine the end behavior. - Evaluate a few useful x-values, including the y-intercept, to refine your description of the graph.

1. For \(n = 1\), \(x = 1\) is a simple zero, so the graph crosses the x-axis there. For \(n = 2\), \(x = 1\) is a double zero, so the graph touches the x-axis at a local minimum. 2. For \(n = 3\), \(f_3(x) = \frac{1}{4}(x + 2)(x - 1)^3\). The graph crosses at the simple zero \(x = -2\) and crosses with flattening at the triple zero \(x = 1\). The polynomial has degree 4 and a positive leading coefficient, so both ends rise. 3. For even \(n\), the graph touches the x-axis at \(x = 1\). For odd \(n\), it crosses there; when \(n \geq 3\), the crossing becomes flatter as the multiplicity increases. As \(n\) increases, the graph becomes flatter near \(x = 1\).

a) For \(n = 1\), the graph crosses the x-axis at \(x = 1\). For \(n = 2\), it touches the x-axis there at a local minimum. b) For \(n = 3\), the graph has a simple zero at \(x = -2\), a triple zero where it crosses with flattening at \(x = 1\), and both ends rise. c) Even values of \(n\) produce a touch at \(x = 1\), while odd values produce a crossing. Larger values of \(n\) make the graph flatter near the zero.

- Factor out the greatest power of \(x\). - Use the exponent on each factor to determine multiplicity. - Compare the exponents in each expanded function to the definitions of even and odd functions.

1. Factor the family: \(g_n(x) = \frac{1}{10}x^{n-2}(x^2 - 16) = \frac{1}{10}x^{n-2}(x - 4)(x + 4)\). 2. For \(n = 3\), \(g_3(x) = \frac{1}{10}x(x - 4)(x + 4)\). The zeros \(-4\), \(0\), and \(4\) all have multiplicity \(1\). 3. For \(n = 4\), \(g_4(x) = \frac{1}{10}x^2(x - 4)(x + 4)\). The zero \(0\) has multiplicity \(2\), while \(-4\) and \(4\) each have multiplicity \(1\). 4. The function \(g_3(x) = \frac{1}{10}(x^3 - 16x)\) contains only odd powers, so it is odd. 5. The function \(g_4(x) = \frac{1}{10}(x^4 - 16x^2)\) contains only even powers, so it is even. 6. The function \(g_5(x) = \frac{1}{10}(x^5 - 16x^3)\) contains only odd powers, so it is odd and its graph is symmetric about the origin.

a) \(g_3\): \(-4\), \(0\), and \(4\), each with multiplicity \(1\). \(g_4\): \(0\) with multiplicity \(2\), and \(-4\) and \(4\) with multiplicity \(1\). b) \(g_3\) is odd; \(g_4\) is even. c) \(g_5\) is odd, so its graph is symmetric about the origin.

- Read the three x-intercepts from the graph. - Use each zero to write a factor, paying attention to its sign. - Substitute the coordinates of \(P\) to solve for \(a\).

1. The graph crosses the x-axis at \(x = -4\), \(x = 0\), and \(x = 4\). These are three simple zeros. 2. Write \(f(x) = a(x + 4)x(x - 4)\). 3. Substitute \(P(2, 2.4)\): \(2.4 = a(2 + 4)(2)(2 - 4)\). 4. Thus, \(2.4 = -24a\), so \(a = -0.1\). 5. Therefore, \(f(x) = -0.1x(x + 4)(x - 4)\).

\(f(x) = -0.1x(x + 4)(x - 4)\)

- Set the two function expressions equal. - Try factoring the resulting polynomial by grouping. - Relate a repeated solution of \(f(x) - g(x) = 0\) to whether the graphs cross or touch.

1. Set the functions equal: \(x^3 - x = x^2 - 1\). 2. Move all terms to one side: \(x^3 - x^2 - x + 1 = 0\). 3. Factor by grouping: \(x^2(x - 1) - (x - 1) = (x^2 - 1)(x - 1) = (x - 1)^2(x + 1)\). 4. The solutions are \(x = 1\) with multiplicity 2 and \(x = -1\) with multiplicity 1. 5. Substitution gives \(f(1) = g(1) = 0\) and \(f(-1) = g(-1) = 0\). 6. Therefore, the intersection points are \((1, 0)\) and \((-1, 0)\). The double solution at \(x = 1\) means the graphs touch there rather than cross.

The intersection points are \((-1, 0)\) and \((1, 0)\). At \(x = 1\), the graphs touch.

- Identify where each graph crosses or touches the x-axis. - A touch indicates an even multiplicity. - A crossing usually indicates an odd multiplicity. - Use the y-intercept as an additional check.

1. Equation (A) has a double zero at \(x = -2\) and a simple zero at \(x = 1\). Its graph touches at \(x = -2\) and crosses at \(x = 1\), so (A) matches \(f\). 2. Equation (B) has a simple zero at \(x = -2\) and a double zero at \(x = 1\). Its negative leading coefficient makes the right end fall, so (B) matches \(g\). 3. Equation (C) has three simple zeros at \(x = -2\), \(x = 0\), and \(x = 2\). Its graph crosses the x-axis at all three zeros, so (C) matches \(h\).

(A) \(f\) (B) \(g\) (C) \(h\)

- Compare the degree and leading coefficient with the graph's end behavior. - Determine the multiplicity of each zero from the factored equation. - Compare every x-intercept of the displayed graph with the zeros of \(g\).

1. Function \(g\) is a cubic with a negative leading coefficient, so its right end must fall. The displayed graph has both ends rising. 2. Function \(g\) has a simple zero at \(x = -1\), so it should cross the x-axis there. The displayed graph touches the x-axis at \(x = -1\). 3. Function \(g\) has a double zero at \(x = 2\), so it should touch the x-axis there. The displayed graph crosses the x-axis at \(x = 2\). 4. The displayed graph also has an additional zero at \(x = -2\), which \(g\) does not have.

Any three of the following: 1. The displayed graph rises to the right, but \(g\) must fall to the right. 2. The displayed graph touches at \(x = -1\), but \(g\) must cross there. 3. The displayed graph crosses at \(x = 2\), but \(g\) must touch there. 4. The displayed graph has a zero at \(x = -2\), but \(g\) does not.

- Read the three x-intercepts from the graph. - Use the zeros to write a factored equation with an unknown leading factor. - Use the y-intercept to determine the leading factor. - Check the resulting equation against the graph's end behavior.

1. The graph crosses the x-axis at \(x = -1\), \(x = 1\), and \(x = 4\). 2. Write \(g(x) = a(x + 1)(x - 1)(x - 4)\). 3. The y-intercept is \((0, 2)\). Substitute it: \(2 = a(1)(-1)(-4) = 4a\). 4. Thus, \(a = 0.5\). 5. Therefore, \(g(x) = 0.5(x + 1)(x - 1)(x - 4)\). Expanding gives \(g(x) = 0.5x^3 - 2x^2 - 0.5x + 2\).

\(g(x) = 0.5(x + 1)(x - 1)(x - 4)\)

- Use crossing and touching behavior to determine the exponent of each factor. - Check that the multiplicities add to the degree of the polynomial. - Substitute the coordinates of \(P\) to determine the leading factor.

1. The graph touches the x-axis at \(x = -2\), so \(x = -2\) is a double zero. It crosses at \(x = 1\) and \(x = 4\), so those are simple zeros. 2. Write \(h(x) = a(x + 2)^2(x - 1)(x - 4)\). 3. Use \(P(3, 5)\): \(5 = a(3 + 2)^2(3 - 1)(3 - 4)\). 4. Thus, \(5 = -50a\), so \(a = -0.1\). 5. Therefore, \(h(x) = -0.1(x + 2)^2(x - 1)(x - 4)\).

\(h(x) = -0.1(x + 2)^2(x - 1)(x - 4)\)

- Compare a simple zero with a double zero on the graphs. - Determine which algebraic form must produce y-axis symmetry. - Use each y-intercept to determine one parameter. - Substitute one additional point to determine the remaining parameter.

1. Graph a crosses the x-axis at \(x = -2\) and touches it at \(x = 1\). This matches the simple factor \((x + 2)\) and the squared factor \((x - 1)^2\), so graph a represents \(f\). 2. Graph b is symmetric about the y-axis, which matches the even function \(g(x) = bx^4 + c\). 3. Graph a has y-intercept \(1\). Therefore, \(1 = a(0 + 2)(0 - 1)^2 = 2a\), so \(a = 0.5\). 4. Graph b has y-intercept \(2\), so \(c = 2\). 5. Graph b passes through \((2, 0)\). Thus, \(0 = b(2)^4 + 2 = 16b + 2\), so \(b = -\frac{1}{8} = -0.125\).

a) Graph a represents \(f\), and graph b represents \(g\). b) \(a = 0.5\), \(b = -\frac{1}{8}\), and \(c = 2\)

- Determine whether each x-intercept is a crossing or a touch. - Use the smallest exponent consistent with each graph behavior. - A touch indicates an even multiplicity; a crossing indicates an odd multiplicity. - Choose a point that lies exactly on a grid intersection to determine the leading factor.

1. For Graph a), the graph crosses at \(x = -3\) and \(x = 2\) and touches at \(x = 0\). Thus, \(f(x) = ax^2(x + 3)(x - 2)\). 2. The graph passes through \((-2, 4)\). Substitute: \(4 = a(-2)^2(-2 + 3)(-2 - 2) = -16a\), so \(a = -0.25\). Therefore, \(f(x) = -0.25x^2(x + 3)(x - 2)\). 3. For Graph b), the graph crosses at \(x = -2\) and \(x = 1\) and touches at \(x = 3\). Thus, \(g(x) = a(x + 2)(x - 1)(x - 3)^2\). 4. The graph passes through \((2, 1)\). Substitute: \(1 = a(2 + 2)(2 - 1)(2 - 3)^2 = 4a\), so \(a = 0.25\). Therefore, \(g(x) = 0.25(x + 2)(x - 1)(x - 3)^2\).

a) \(f(x) = -0.25x^2(x + 3)(x - 2)\) b) \(g(x) = 0.25(x + 2)(x - 1)(x - 3)^2\)

- Read each x-intercept from the graph. - Decide whether the graph crosses or touches at each zero. - Use the zeros to write a factored equation with an unknown leading factor. - Substitute the marked point to determine the leading factor. - Expand the factors one pair at a time.

1. The graph crosses the x-axis at \(x = -3\), \(x = -1\), and \(x = 2\), so all three zeros have multiplicity 1. 2. Write \(f(x) = a(x + 3)(x + 1)(x - 2)\). 3. Substitute \(P(0, -3)\): \(-3 = a(3)(1)(-2) = -6a\), so \(a = 0.5\). Thus, \(f(x) = 0.5(x + 3)(x + 1)(x - 2)\). 4. Expand: \((x + 3)(x + 1) = x^2 + 4x + 3\), and \((x^2 + 4x + 3)(x - 2) = x^3 + 2x^2 - 5x - 6\). Therefore, \(f(x) = 0.5x^3 + x^2 - 2.5x - 3\).

a) \(x = -3\), \(x = -1\), and \(x = 2\), each with multiplicity 1 b) \(f(x) = 0.5(x + 3)(x + 1)(x - 2)\) c) \(f(x) = 0.5x^3 + x^2 - 2.5x - 3\)

- Distinguish a crossing from a touch at each x-intercept. - A crossing indicates an odd multiplicity, while a touch indicates an even multiplicity. - Represent a double zero with a squared factor. - Substitute the given point to solve for the leading factor.

1. The graph crosses the x-axis at \(x = -2\) and \(x = 3\), so these are simple zeros. It touches the x-axis at \(x = 1\), so \(x = 1\) is a double zero. The multiplicities add to \(1 + 2 + 1 = 4\). 2. Write \(g(x) = a(x + 2)(x - 1)^2(x - 3)\). 3. Substitute \(Q(2, 2)\): \(2 = a(2 + 2)(2 - 1)^2(2 - 3) = -4a\). 4. Thus, \(a = -0.5\), so \(g(x) = -0.5(x + 2)(x - 1)^2(x - 3)\).

a) \(x = -2\) and \(x = 3\) are simple zeros; \(x = 1\) is a double zero. b) \(g(x) = -0.5(x + 2)(x - 1)^2(x - 3)\)

- Use the sign of the leading coefficient to determine whether both ends rise or fall. - Factor each expression to identify the zeros and their multiplicities. - Use touching versus crossing behavior at the x-axis. - Compare the y-intercepts as a final check.

1. Equation \(y_C\) has a negative leading coefficient, so both ends fall. This matches \(k_2\). 2. Factor \(y_A\): \(y_A = (x^2 - 1)^2 = (x - 1)^2(x + 1)^2\). It has double zeros at \(x = -1\) and \(x = 1\), so its graph touches the x-axis at both zeros. This matches \(k_3\). 3. Factor \(y_B\): \(y_B = (x^2 - 1)(x^2 - 4) = (x - 1)(x + 1)(x - 2)(x + 2)\). It has four simple zeros at \(x = -2\), \(x = -1\), \(x = 1\), and \(x = 2\), so its graph crosses at all four zeros. This matches \(k_1\). 4. The y-intercepts provide a check: \(y_A(0) = 1\), \(y_B(0) = 4\), and \(y_C(0) = -2\).

\(k_1\): \(y_B = x^4 - 5x^2 + 4\) \(k_2\): \(y_C = -0.5x^4 + 2.5x^2 - 2\) \(k_3\): \(y_A = x^4 - 2x^2 + 1\)

- A graph that only touches the x-axis has a multiple zero; use the zeros to write a factored form with an unknown leading factor. - Use the y-intercept to determine the factor.

1. The graph has a double zero at \(x=-1\), a simple zero at \(x=2\), and y-intercept \((0, 2)\). 2. Write \(f(x)=k(x+1)^2(x-2)\). Using \(f(0)=2\) gives \(-2k=2\), so \(k=-1\). 3. Expanding, \(f(x)=-(x+1)^2(x-2)=-x^3+3x+2\). Therefore, \(a=-1\), \(b=3\), and \(c=2\).

\(a=-1\), \(b=3\), \(c=2\)

- Use the y-axis symmetry and y-intercept. - Use the four visible zeros to write a factored form.

1. The graph is symmetric about the y-axis, consistent with a polynomial containing only even powers. Its y-intercept is \(4\), so \(c = 4\). 2. The zeros are \(x = -2\), \(x = -1\), \(x = 1\), and \(x = 2\). 3. Write \(f(x) = a(x^2 - 1)(x^2 - 4)\). 4. Use \((0, 4)\): \(4 = a(-1)(-4) = 4a\), so \(a = 1\). 5. Expanding gives \(f(x) = (x^2 - 1)(x^2 - 4) = x^4 - 5x^2 + 4\).

\(f(x) = x^4 - 5x^2 + 4\)

- A touch at the x-axis indicates an even multiplicity. - Simplify \((x - 2)^2(x + 2)^2\) before expanding.

1. The graph touches the x-axis at \(x = -2\) and \(x = 2\), so both zeros have multiplicity 2. The y-intercept is \((0, 4)\). 2. Write \(f(x) = a(x - 2)^2(x + 2)^2 = a(x^2 - 4)^2\). 3. Substitute \((0, 4)\): \(4 = a(0^2 - 4)^2 = 16a\), so \(a = 0.25\). 4. Expand: \(f(x) = 0.25(x^4 - 8x^2 + 16) = 0.25x^4 - 2x^2 + 4\).

\(f(x) = 0.25x^4 - 2x^2 + 4\)

- A flattened crossing at the origin indicates a zero of odd multiplicity greater than 1. - Use the other zero and one additional point to determine the equation.

1. The graph crosses with flattening at \(x = 0\), so \(x = 0\) is a triple zero. It also has a simple zero at \(x = -2\). 2. Write \(f(x) = ax^3(x + 2)\). 3. Use the point \((-1, 1)\): \(1 = a(-1)^3(-1 + 2) = -a\), so \(a = -1\). 4. Therefore, \(f(x) = -x^3(x + 2) = -x^4 - 2x^3\).

\(f(x) = -x^4 - 2x^3\)

- Factor the given form and use how the graph behaves at each x-intercept to determine multiplicity. - Use the other x-intercept to write a factored form. - Use another exact grid point on the curve to determine the leading factor.

1. The given form factors as \(f(x)=x^3(ax+b)\). The graph crosses the x-axis at \(x=0\) with multiplicity \(3\) and has its other zero at \(x=4\). Thus, \(f(x)=k x^3(x-4)\). 2. The graph passes through the grid point \((2, -4)\). Substituting gives \(-4=k\cdot8\cdot(-2)\), so \(k=\frac{1}{4}\). 3. Therefore, \(f(x)=\frac{1}{4}x^4-x^3\), so \(a=\frac{1}{4}\) and \(b=-1\).

\(a=\frac{1}{4}\), \(b=-1\)

- Identify whether each x-intercept is a crossing or a touch. - Use the behavior at each zero to choose the factor's exponent. - Substitute the y-intercept to determine the leading factor.

1. The graph touches the x-axis at \(x = -2\), so \(x = -2\) is a double zero. It crosses at \(x = 2\), so \(x = 2\) is a simple zero. 2. Write \(f(x) = a(x + 2)^2(x - 2)\). 3. The y-intercept is \((0, -2)\). Substitute: \(-2 = a(2)^2(-2) = -8a\), so \(a = 0.25\). 4. Therefore, \(f(x) = 0.25(x + 2)^2(x - 2)\).

\(f(x) = 0.25(x + 2)^2(x - 2)\)

- Use the y-axis symmetry to check the locations of the zeros. - A local extremum at the origin on the x-axis indicates a double zero. - Substitute the coordinates of the marked point into the factored equation.

1. The graph touches the x-axis at \(x = 0\), so \(x = 0\) is a double zero. It crosses at \(x = -3\) and \(x = 3\), so these are simple zeros. 2. Write \(f(x) = ax^2(x + 3)(x - 3) = ax^2(x^2 - 9)\). 3. Substitute \(P(2, -2)\): \(-2 = a(2)^2(2^2 - 9) = -20a\). 4. Thus, \(a = 0.1\), so \(f(x) = 0.1x^2(x^2 - 9)\).

\(f(x) = 0.1x^2(x^2 - 9)\)

- A triple zero at \(x = 1\) gives a factor of \((x - 1)^3\). - Include an unknown leading factor, then substitute the given point.

1. A triple zero at \(x = 1\) gives the form \(f(x) = a(x - 1)^3\). 2. Use the point \(S(3, 2)\): \(2 = a(3 - 1)^3 = 8a\). 3. Thus, \(a = 0.25\). 4. Therefore, \(f(x) = 0.25(x - 1)^3\).

\(f(x) = 0.25(x - 1)^3\)

- Read the x-intercepts from the graph. - Use origin symmetry to check the possible powers of \(x\). - Write the equation in factored form using the zeros. - Substitute the marked point to determine the leading factor.

1. The zeros are \(x = -2\), \(x = 0\), and \(x = 2\). 2. The graph crosses at all three zeros, so the lowest possible degree is 3. Its origin symmetry is consistent with an odd polynomial. 3. Write \(f(x) = ax(x - 2)(x + 2) = a(x^3 - 4x)\). 4. Substitute \(P(1, -1.5)\): \(-1.5 = a(1^3 - 4) = -3a\). 5. Thus, \(a = 0.5\), so \(f(x) = 0.5x^3 - 2x\).

a) \(x = -2\), \(x = 0\), and \(x = 2\) b) \(f(x) = 0.5x^3 - 2x\)

- Relate a simple or double zero to crossing or touching the x-axis. - Use the sign of the leading coefficient to determine the right-end behavior. - Compare the zeros in each equation with the x-intercepts of the graphs. - A touch indicates an even multiplicity.

1. Function \(p\) has a double zero at \(x = 0\), a simple zero at \(x = 3\), and a positive leading coefficient. Its graph touches at \(0\), crosses at \(3\), and rises to the right. This matches Graph A. 2. Function \(q\) has a simple zero at \(x = 0\), a double zero at \(x = 3\), and a positive leading coefficient. Its graph crosses at \(0\), touches at \(3\), and rises to the right. This matches Graph B. 3. Function \(r\) has the same zero multiplicities as \(p\), but its leading coefficient is negative, so its right end falls. This matches Graph C.

(1) Graph A (2) Graph B (3) Graph C

- Use the exponent on each factor to determine whether the graph should cross or touch. - Compare the graph behavior at both zeros. - Evaluate the equation at \(x = 0\). - Check whether the displayed graph has any other relevant features.

1. In \(h\), the factor \((x - 1)^2\) gives a double zero at \(x = 1\), so the graph of \(h\) should touch the x-axis there. The displayed graph crosses there. 2. In \(h\), the factor \((x + 2)\) gives a simple zero at \(x = -2\), so the graph of \(h\) should cross there. The displayed graph touches there. 3. The y-intercept of \(h\) is \(h(0) = (-1)^2(2) = 2\), but the displayed graph has y-intercept \(-4\).

The claim is false because: 1. \(h\) should touch at \(x = 1\), but the displayed graph crosses there. 2. \(h\) should cross at \(x = -2\), but the displayed graph touches there. 3. \(h\) has y-intercept \(2\), but the displayed graph has y-intercept \(-4\).

- Count the turning points and inspect the graph's end behavior. - Use the \(180^\circ\) rotational symmetry about the origin. - Write the polynomial in factored form from its zeros. - Substitute the additional point to find the leading factor.

1. The graph has two turning points, so its degree is at least \(3\). A cubic is the least possible degree and is consistent with the graph's end behavior. 2. The graph has \(180^\circ\) rotational symmetry about the origin, so \(g\) is odd. A general cubic with this symmetry is \(g(x)=ax^3+cx\). 3. Using the three zeros, write \(g(x)=a x(x+3)(x-3)=a(x^3-9x)\). Since \(P(1, -2)\) lies on the graph, \(-2=a(1-9)=-8a\), so \(a=\frac{1}{4}\). Therefore, \(g(x)=\frac{1}{4}x^3-\frac{9}{4}x\).

1. Least possible degree: \(3\) 2. \(g(x)=ax^3+cx\) 3. \(g(x)=\frac{1}{4}x^3-\frac{9}{4}x\)

- Distinguish a crossing from a touch at each x-intercept. - Represent a touching zero with a squared factor. - Substitute the marked point to determine the leading factor.

1. The graph crosses the x-axis at \(x = 0\), so \(x = 0\) is a simple zero. It touches at \(x = 4\), so \(x = 4\) is a double zero. 2. Write \(f(x) = ax(x - 4)^2\). 3. Use \(Q(2, 2)\): \(2 = a(2)(2 - 4)^2 = 8a\). 4. Thus, \(a = 0.25\). 5. Therefore, \(f(x) = 0.25x(x - 4)^2\).

\(f(x) = 0.25x(x - 4)^2\)

- Use the known zeros to write a factored equation. - A cubic polynomial can have at most three real zeros. - Substitute the given point to determine the leading factor. - Pay close attention to signs when expanding.

1. Use the three zeros to write \(f(x) = a(x + 4)x(x - 2)\). 2. Substitute \(P(-2, 8)\): \(8 = a(-2 + 4)(-2)(-2 - 2)\). 3. Thus, \(8 = 16a\), so \(a = 0.5\). 4. The factored form is \(f(x) = 0.5x(x + 4)(x - 2)\). 5. Expanding gives \(f(x) = 0.5x^3 + x^2 - 4x\).

\(f(x) = 0.5x^3 + x^2 - 4x\)

- Use the discriminant to determine when real zeros exist. - The vertex lies halfway between the zeros. - Expand \(a(x - x_1)(x - x_2)\) and compare the coefficient of \(x\).

1. Real zeros exist when the discriminant is nonnegative: \(D = k^2 - 4 \cdot 2 \cdot 8 = k^2 - 64 \ge 0\). 2. Solving \(k^2 \ge 64\) gives \(k \le -8\) or \(k \ge 8\). 3. The x-coordinate of the vertex is \(x = -\frac{k}{2 \cdot 2} = -\frac{k}{4}\). As a positive \(k\) becomes large, the midpoint of the zeros moves farther left, not closer to the origin. The claim is false. 4. If the zeros are \(x_1\) and \(x_2\), then \(g(x) = a(x - x_1)(x - x_2)\). 5. Expanding gives \(g(x) = ax^2 - a(x_1 + x_2)x + ax_1x_2\). Therefore, \(b = -a(x_1 + x_2)\). 6. If \(x_1 = -x_2\), then \(x_1 + x_2 = 0\), so \(b = 0\).

a) \(k \le -8\) or \(k \ge 8\) b) The claim is false because the midpoint of the zeros is \(-\frac{k}{4}\), which moves away from the origin as positive \(k\) increases. c) Opposite zeros have sum \(0\), and coefficient comparison gives \(b = -a(x_1 + x_2) = 0\).

- Factor the polynomial to identify one zero that is present for every value of \(k\). - Use the discriminant to determine how many real zeros the quadratic factor has. - Check when the quadratic factor produces the zero that is already present. - Organize the result by the three possible signs of the discriminant.

1. Factor out \(x\): \(f_k(x) = x(x^2 - 4x + k)\). Therefore, \(x = 0\) is always a zero. 2. The discriminant of \(x^2 - 4x + k\) is \(D = (-4)^2 - 4(1)(k) = 16 - 4k\). 3. If \(k > 4\), then \(D < 0\), so the quadratic factor has no real zeros. The function has only the zero \(x = 0\). 4. If \(k = 4\), then \(D = 0\), and the quadratic factor has the repeated zero \(x = 2\). Together with \(x = 0\), the function has two distinct real zeros. 5. If \(k < 4\), then the quadratic factor has two distinct real zeros. One of these equals the existing zero \(x = 0\) only when \(k = 0\). For \(k = 0\), \(f_0(x) = x^2(x - 4)\), so there are two distinct real zeros. 6. For \(k < 4\) with \(k \ne 0\), the two quadratic zeros are distinct from \(0\), so the function has three distinct real zeros.

a) \(k > 4\) b) \(k = 0\) or \(k = 4\) c) \(k < 4\) and \(k \ne 0\)

- For \(x = 2\) to be a repeated zero, both factors must be zero there. - A quadratic has a repeated real zero when its discriminant equals zero. - After finding \(b\), factor the quadratic to identify the repeated zero. - For a triple zero, all three linear factors would need to correspond to the same value of \(x\).

1. For \(x = 2\) to be a double zero, the quadratic factor must also equal zero at \(x = 2\): \(2^2 + 2b + 25 = 0\). Thus, \(2b + 29 = 0\), so \(b = -\frac{29}{2} = -14.5\). 2. A double zero different from \(2\) occurs when the quadratic factor has one repeated real zero. Its discriminant must be zero: \(b^2 - 4(1)(25) = b^2 - 100 = 0\). Therefore, \(b = -10\) or \(b = 10\). 3. If \(b = -10\), then the quadratic is \((x - 5)^2\), so the double zero is \(x = 5\). If \(b = 10\), then the quadratic is \((x + 5)^2\), so the double zero is \(x = -5\). 4. A triple zero would require the quadratic factor to be \((x - 2)^2 = x^2 - 4x + 4\). Its constant term would have to be \(4\), but the given constant term is fixed at \(25\). Therefore, a triple zero is impossible.

a) \(b = -14.5\) b) \(b = -10\) or \(b = 10\) c) A triple zero would require \(x^2 + bx + 25 = (x - 2)^2\), but the constant terms \(25\) and \(4\) cannot be equal.

- Y-axis symmetry means zeros occur in opposite pairs. - A graph that touches the x-axis has a zero of even multiplicity. - Use the degree to determine the multiplicities. - Include a nonzero scale factor to describe the full family.

1. Y-axis symmetry means the polynomial is even. Therefore, if \(x=2\) is a zero, then \(x=-2\) is also a zero. 2. A local extremum on the x-axis means the graph touches, rather than crosses, the axis. Thus, each zero has even multiplicity. 3. Because the polynomial has degree \(4\), both zeros must have multiplicity \(2\). Therefore, \(f_a(x)=a(x-2)^2(x+2)^2\), where \(a\ne0\). 4. Equivalently, \(f_a(x)=a(x^2-4)^2=ax^4-8ax^2+16a\).

\(f_a(x)=a(x^2-4)^2\), where \(a\in\mathbb R\setminus\{0\}\)

- Apply the difference-of-squares identity more than once when useful. - Squaring an expression removes an overall negative sign. - Check whether a factor such as \(x^2 + 1\) has real zeros. - Read multiplicity from the exponent on each linear factor.

1. At \(x = 1\), all five functions equal \(0\). At \(x = 0\), the values are \(g_1(0) = 1\), \(g_2(0) = -1\), \(g_3(0) = 1\), \(g_4(0) = 1\), and \(g_5(0) = -1\). 2. Factor \(g_1\): \((x^2 - 1)^2 = (x - 1)^2(x + 1)^2\). Because squaring removes a negative sign, \((1 - x^2)^2 = (x^2 - 1)^2\). Therefore, \(g_1\), \(g_3\), and \(g_4\) each have zeros \(x = -1\) and \(x = 1\), both with multiplicity 2. 3. Factor \(g_2\): \(x^4 - 1 = (x^2 - 1)(x^2 + 1) = (x - 1)(x + 1)(x^2 + 1)\). Thus, \(g_2 = g_5\). Since \(x^2 + 1\) has no real zeros, both functions have zeros \(x = -1\) and \(x = 1\), each with multiplicity 1. 4. Therefore, \(g_1 = g_3 = g_4\), and \(g_2 = g_5\).

a) At \(x = 1\), all five values are \(0\). At \(x = 0\): \(g_1 = 1\), \(g_2 = -1\), \(g_3 = 1\), \(g_4 = 1\), \(g_5 = -1\). b) \(g_1\), \(g_3\), and \(g_4\): \(x = -1\) and \(x = 1\), each with multiplicity 2 \(g_2\) and \(g_5\): \(x = -1\) and \(x = 1\), each with multiplicity 1 c) \(g_1 = g_3 = g_4\), and \(g_2 = g_5\).

- Use the difference-of-squares identity before expanding everything. - Check whether the higher-degree terms cancel. - Factor the simplified quadratic. - A squared factor has even multiplicity and does not change sign.

1. Use the difference-of-squares identity: \((x^2-3)(x^2+3)=x^4-9\). 2. Then \(x^4-(x^4-9)=9\). 3. Expand \((x-5)^2=x^2-10x+25\) and combine all terms: \(h(x)=9+x^2-10x+25-9=x^2-10x+25\). Thus, \(h\) is quadratic. 4. Factor: \(h(x)=(x-5)^2\). The only zero is \(x=5\), with multiplicity \(2\). 5. A zero of even multiplicity does not produce a sign change. Therefore, \(h\) does not change sign at any x-value.

\(h(x)=x^2-10x+25=(x-5)^2\). It has no sign-change points.

- Y-axis symmetry means the function is even. - Use multiplicities so their sum equals the required degree. - An odd-degree polynomial must have at least one real zero. - To avoid real zeros, choose an expression that stays strictly positive or strictly negative.

1. For a), y-axis symmetry requires an even function. The factor \(x^2 - 9\) gives the zeros \(x = -3\) and \(x = 3\). Squaring it produces degree \(4\) without adding zeros: \(f(x) = (x^2 - 9)^2\). 2. For b), the multiplicities must add to \(5\). One choice is to make \(x = 0\) a simple zero and \(x = -2\) and \(x = 2\) double zeros: \(f(x) = x(x + 2)^2(x - 2)^2\). 3. For c), \(x^4 \ge 0\) for every real \(x\), so \(x^4 + 1 > 0\). Therefore, \(f(x) = x^4 + 1\) has degree \(4\) and no real zeros.

a) \(f(x) = (x^2 - 9)^2\) b) \(f(x) = x(x + 2)^2(x - 2)^2\) c) \(f(x) = x^4 + 1\)

- Substitute the x-coordinate of a point to check whether it lies on every graph. - Evaluate \((x - 1)^n\) at \(x = 1\) and \(x = 2\). - To rule out other common points, compare two members of the family with different exponents. - Relate odd and even multiplicities to crossing and touching behavior. - Near \(x = 1\), compare \(u^4\) and \(u^2\) for small values of \(|u|\).

1. Substitute each x-coordinate. For every positive integer \(n\), \(f_n(-1) = 0\), \(f_n(1) = 0\), and \(f_n(2) = 3\). Therefore, all graphs pass through \(P\), \(Q\), and \(R\). 2. Any point common to every graph must, in particular, lie on both \(f_1\) and \(f_2\). Solve \((x + 1)(x - 1) = (x + 1)(x - 1)^2\). This simplifies to \((x + 1)(x - 1)(x - 2) = 0\), so \(x = -1\), \(x = 1\), or \(x = 2\). These are exactly the three common points already found. 3. At \(x = 1\), the multiplicity is \(n\). For \(n = 1\), the graph crosses the x-axis with nonzero slope. For \(n = 2\), it touches the x-axis and does not change sign. For \(n = 3\), it crosses the x-axis with a horizontal tangent. 4. In general, a zero of odd multiplicity produces a sign change, while a zero of even multiplicity does not. Multiplicity greater than 1 produces a horizontal tangent at the zero. 5. For \(f_4\), \(x = 1\) is a zero of multiplicity 4, so the graph touches the x-axis without changing sign. Near \(x = 1\), \((x - 1)^4\) is smaller in magnitude than \((x - 1)^2\), so \(f_4\) stays closer to the x-axis and has flatter contact than \(f_2\).

a) Every graph contains \(P(-1, 0)\), \(Q(1, 0)\), and \(R(2, 3)\). Comparing \(f_1\) and \(f_2\) shows that no other x-values can be common to the entire family. b) At \(x = 1\), \(f_1\) crosses with nonzero slope, \(f_2\) touches without crossing, and \(f_3\) crosses with a horizontal tangent. Odd multiplicity gives a sign change; even multiplicity does not. c) The zero \(x = 1\) has multiplicity 4, so there is no sign change. Near the zero, \(f_4\) has flatter contact with the x-axis than \(f_2\).

- A flattened crossing at an x-intercept indicates multiplicity 3 in this problem. - Use the end behavior to check the sign of the leading factor. - The y-intercept is often a convenient point for determining the leading factor.

1. For Graph a), the graph touches at \(x = -1\) and crosses at \(x = 1\) and \(x = 3\). Write \(h(x) = a(x + 1)^2(x - 1)(x - 3)\). 2. The graph passes through \((0, 1)\). Thus, \(1 = a(1)^2(-1)(-3) = 3a\), so \(a = \frac{1}{3}\). Therefore, \(h(x) = \frac{1}{3}(x + 1)^2(x - 1)(x - 3)\). 3. For Graph b), the graph crosses with flattening at \(x = -2\), indicating a triple zero, and crosses normally at \(x = 2\), indicating a simple zero. Write \(p(x) = a(x + 2)^3(x - 2)\). 4. The graph passes through \((0, 4)\). Thus, \(4 = a(2)^3(-2) = -16a\), so \(a = -0.25\). Therefore, \(p(x) = -0.25(x + 2)^3(x - 2)\).

a) \(h(x) = \frac{1}{3}(x + 1)^2(x - 1)(x - 3)\) b) \(p(x) = -0.25(x + 2)^3(x - 2)\)

- Examine how the graph meets the x-axis at each zero. - A graph that turns around at the x-axis has an even multiplicity. - A normal crossing has an odd multiplicity, usually 1. - A flattened crossing has an odd multiplicity of at least 3. - The degree must be at least the sum of the zero multiplicities.

1. The zeros are \((-2, 0)\), \((0, 0)\), and \((2, 0)\). 2. At \(x = -2\), the graph touches the x-axis and turns around, so the minimum multiplicity is 2. 3. At \(x = 0\), the graph crosses the x-axis without flattening, so the minimum multiplicity is 1. 4. At \(x = 2\), the graph crosses the x-axis with flattening, so the minimum multiplicity is 3. 5. The minimum possible degree is the sum of these minimum multiplicities: \(2 + 1 + 3 = 6\).

a) \((-2, 0)\): touch; \((0, 0)\): crossing; \((2, 0)\): flattened crossing b) Minimum multiplicities: \(2\), \(1\), and \(3\), respectively c) Minimum degree: \(6\)

- Read the intersections of \(f\) and \(j\) from the graph. - Intersections occur where two function values are equal. - A product is zero when at least one factor is zero.

1. The intersections of \(f\) and \(j\) occur at \(x=2\) and approximately \(x=-2.8\). 2. Set the formulas for \(g\) and \(j\) equal: \(-x+2=-x^2+4\). 3. Function \(g\) has zero \(2\), and function \(h\) has zero \(-1\). Their product has both zeros: \(m(x)=g(x)h(x)=(-x+2)(x+1)\).

a) \(x=2\) and \(x\approx-2.8\) b) \(-x+2=-x^2+4\) c) One answer is \(m(x)=(-x+2)(x+1)\).