Free Math Worksheets

- A vertex on the x-axis corresponds to one repeated real zero. - Calculate \(b^2 - 4ac\) for each function. - Compare each discriminant with \(0\).

1. A quadratic has its vertex on the x-axis exactly when it has one repeated real zero, so its discriminant is \(0\). 2. For \(f\), \(D = 7^2 - 4 \cdot 1 \cdot 12.25 = 49 - 49 = 0\). Its vertex is on the x-axis. 3. For \(g\), \(D = (-2.4)^2 - 4 \cdot 1 \cdot 1.44 = 5.76 - 5.76 = 0\). Its vertex is on the x-axis. 4. For \(h\), \(D = 1^2 - 4 \cdot 1 \cdot 1 = -3\). Its vertex is not on the x-axis.

a) Yes b) Yes c) No

- Use the same discriminant test for all three functions. - Work with exact fractions. - A zero discriminant means one repeated x-intercept.

1. A parabola touches the x-axis at exactly one point when its discriminant is \(0\). 2. For \(f\), \(D = \left(\frac{2}{3}\right)^2 - 4 \cdot 1 \cdot \left(\frac{1}{9}\right) = 0\). 3. For \(g\), \(D = \left(-\frac{4}{5}\right)^2 - 4 \cdot 1 \cdot \left(\frac{4}{25}\right) = 0\). 4. For \(h\), \(D = \left(\frac{1}{2}\right)^2 - 4 \cdot 1 \cdot \left(\frac{1}{2}\right) = -\frac{7}{4}\). 5. Therefore, \(f\) and \(g\) touch the x-axis once, while \(h\) has no real x-intercepts.

a) \(f\) touches the x-axis at exactly one point. b) \(g\) touches the x-axis at exactly one point. c) \(h\) does not; it has no real x-intercepts.

- Calculate \(b^2 - 4ac\) for each function. - A zero discriminant means the graph has one repeated x-intercept. - Compare the result for each function with \(0\).

1. A parabola has its vertex on the x-axis exactly when the discriminant is \(0\). 2. For \(f\), \(D = 10^2 - 4 \cdot 1 \cdot 25 = 0\), so the vertex is on the x-axis. 3. For \(g\), \(D = (-4)^2 - 4 \cdot 1 \cdot 3 = 4\), so the vertex is not on the x-axis. 4. For \(h\), \(D = 5^2 - 4 \cdot 1 \cdot 6.25 = 0\), so the vertex is on the x-axis.

a) Yes b) No c) Yes

- Recall what a zero discriminant means. - Calculate \(b^2 - 4ac\) for each function. - Compare each result with \(0\).

1. A parabola touches the x-axis at exactly one point when its discriminant is \(0\). 2. For \(f\), \(D = 12^2 - 4 \cdot 1 \cdot 36 = 144 - 144 = 0\), so the graph touches the x-axis once. 3. For \(g\), \(D = (-7)^2 - 4 \cdot 1 \cdot 12 = 49 - 48 = 1\), so the graph crosses the x-axis at two points. 4. For \(h\), \(D = (-1)^2 - 4 \cdot 1 \cdot 0.25 = 1 - 1 = 0\), so the graph touches the x-axis once.

a) \(f\) touches the x-axis at exactly one point. b) \(g\) does not; it crosses the x-axis at two points. c) \(h\) touches the x-axis at exactly one point.

- Identify \(a\), \(b\), and \(c\) carefully for each function. - Calculate \(b^2 - 4ac\). - Use only the sign of the discriminant; the zeros themselves are not required.

1. For \(f\), \(D = (-6)^2 - 4 \cdot 1 \cdot 9 = 36 - 36 = 0\), so there is one real zero. 2. For \(g\), \(D = 2^2 - 4 \cdot 0.5 \cdot 5 = 4 - 10 = -6\), so there are no real zeros. 3. For \(h\), \(D = 4^2 - 4 \cdot (-1) \cdot (-1) = 16 - 4 = 12\), so there are two real zeros.

a) One real zero b) No real zeros c) Two real zeros

- Calculate or identify the discriminant in each case. - A positive, zero, or negative discriminant gives a different number of real solutions. - The actual solution values are not needed.

1. For a), \(D = (-10)^2 - 4 \cdot 1 \cdot 25 = 100 - 100 = 0\), so there is one real solution. 2. For b), \(D = 4^2 - 4 \cdot (-2) \cdot (-5) = 16 - 40 = -24\), so there are no real solutions. 3. For c), \(D = b^2 - 4ac = 49 - 50 = -1\), so there are no real solutions. 4. For d), \(D = 12.5 > 0\), so there are two distinct real solutions.

a) One real solution b) No real solutions c) No real solutions d) Two distinct real solutions

- Identify the three coefficients. - Substitute them into \(b^2 - 4ac\). - Interpret the sign of the result.

1. The coefficients are \(a = \frac{1}{2}\), \(b = 3\), and \(c = 5\). 2. The discriminant is \(D = 3^2 - 4 \cdot \frac{1}{2} \cdot 5 = 9 - 10 = -1\). 3. Since \(D < 0\), the equation has no real solutions.

The discriminant is \(-1\), so there are no real solutions.

- Count the x-intercepts of each graph. - Relate the number of real zeros to the sign of the discriminant.

1. The graph of \(f\) crosses the x-axis twice, so \(f(x) = 0\) has two distinct real solutions. Therefore, \(D > 0\). 2. The graph of \(g\) has no x-intercepts, so \(g(x) = 0\) has no real solutions. Therefore, \(D < 0\).

For \(f\), \(D > 0\). For \(g\), \(D < 0\).

- Use the condition for exactly one real solution. - Identify \(a\), \(b\), and \(c\) in terms of \(k\). - Set the discriminant equal to \(0\) and solve.

1. A quadratic equation has exactly one real solution when its discriminant is \(0\). 2. Here, \(a = k\), \(b = 12\), and \(c = 18\), so \(D = 12^2 - 4 \cdot k \cdot 18 = 144 - 72k\). 3. Set the discriminant equal to zero: \(144 - 72k = 0\). 4. Solve: \(k = 2\).

\(k = 2\)

- Relate touching the x-axis once to the number of real zeros. - Write the discriminant in terms of \(a\). - Factor the resulting equation in \(a\).

1. A quadratic graph touches the x-axis at exactly one point when its discriminant is \(0\). 2. For \(x^2 + ax + a = 0\), the discriminant is \(D = a^2 - 4 \cdot 1 \cdot a = a^2 - 4a\). 3. Set the discriminant equal to zero: \(a^2 - 4a = 0\). 4. Factor: \(a(a - 4) = 0\). 5. Therefore, \(a = 0\) or \(a = 4\).

\(a = 0\) or \(a = 4\)

- Calculate \(b^2 - 4ac\) in terms of \(q\). - Match positive, zero, and negative discriminants with the possible numbers of real solutions. - Solve each resulting inequality or equation for \(q\).

1. For \(x^2 + 6x + q = 0\), the discriminant is \(D = 6^2 - 4 \cdot 1 \cdot q = 36 - 4q = 4(9 - q)\). 2. For exactly two real solutions, require \(D > 0\): \(4(9 - q) > 0\), so \(q < 9\). 3. For exactly one real solution, require \(D = 0\): \(4(9 - q) = 0\), so \(q = 9\). 4. For no real solutions, require \(D < 0\): \(4(9 - q) < 0\), so \(q > 9\).

a) \(q < 9\) b) \(q = 9\) c) \(q > 9\)

- Write each equation in standard form before identifying \(a\), \(b\), and \(c\). - Use the sign of the discriminant to determine the number of real solutions. - After classifying an equation, solve it by the most efficient method.

1. For a), \(D = 0^2 - 4 \cdot 1 \cdot 25 = -100\). Since \(D < 0\), there are no real solutions. 2. For b), expand to \(x^2 - 2\sqrt{2}x + 2 = 0\). Then \(D = (-2\sqrt{2})^2 - 4 \cdot 1 \cdot 2 = 8 - 8 = 0\), so there is one real solution. From \(x - \sqrt{2} = 0\), the solution is \(x = \sqrt{2}\). 3. For c), \(D = 0^2 - 4 \cdot 3 \cdot (-12) = 144\), so there are two real solutions. Solving \(3x^2 = 12\) gives \(x^2 = 4\), so \(x = -2\) or \(x = 2\).

a) No real solutions b) One real solution: \(x = \sqrt{2}\) c) Two real solutions: \(x = -2\) and \(x = 2\)

- Rewrite \(x^2 = a\) in standard form and express its discriminant in terms of \(a\). - Consider separately when \(a\) is negative, zero, or positive. - Apply the same discriminant test to \(x^2 + c = 0\).

1. Rewrite \(x^2 = a\) as \(x^2 - a = 0\). Its discriminant is \(D = 0^2 - 4 \cdot 1 \cdot (-a) = 4a\). 2. If \(a < 0\), then \(D < 0\), so there are no real solutions. For example, \(x^2 = -4\) has no real solutions. 3. If \(a = 0\), then \(D = 0\), so there is exactly one real solution, \(x = 0\). 4. If \(a > 0\), then \(D > 0\), so there are two real solutions. For example, \(x^2 = 9\) has \(x = -3\) and \(x = 3\). 5. The zeros of \(f(x) = x^2 + c\) satisfy \(x^2 + c = 0\), whose discriminant is \(D = -4c\). There are no real zeros when \(-4c < 0\), which is equivalent to \(c > 0\).

The claim is false. If \(a < 0\), there are no real solutions; if \(a = 0\), there is one real solution; and if \(a > 0\), there are two real solutions. The function \(f(x) = x^2 + c\) has no real zeros when \(c > 0\).

- What must the discriminant equal when a quadratic has exactly one real zero? - Why does a vertex on the \(x\)-axis correspond to exactly one real zero? - After finding \(k\), can you rewrite each quadratic as a perfect square?

1. If the vertex lies on the \(x\)-axis, the quadratic has exactly one real zero, so its discriminant is \(0\). 2. Set the discriminant equal to zero: \(k^2 - 4 \cdot 1 \cdot 9 = 0\), so \(k^2 - 36 = 0\). 3. Thus, \(k = 6\) or \(k = -6\). 4. For \(k = 6\), \(f_6(x) = x^2 + 6x + 9 = (x + 3)^2\), so the vertex is \((-3, 0)\). For \(k = -6\), \(f_{-6}(x) = x^2 - 6x + 9 = (x - 3)^2\), so the vertex is \((3, 0)\).

\(k = 6\) gives vertex \((-3, 0)\), and \(k = -6\) gives vertex \((3, 0)\).

- Write the discriminant in terms of \(c\). - Set the discriminant equal to \(0\) for exactly one real solution. - Substitute \(c = 20\) into the same discriminant expression.

1. For \(x^2 - 8x + c = 0\), the discriminant is \(D = (-8)^2 - 4 \cdot 1 \cdot c = 64 - 4c\). 2. For exactly one real solution, set \(D = 0\): \(64 - 4c = 0\), so \(c = 16\). 3. When \(c = 16\), the equation is \(x^2 - 8x + 16 = 0\), or \((x - 4)^2 = 0\). Thus, the repeated solution is \(x = 4\). 4. When \(c = 20\), \(D = 64 - 4 \cdot 20 = -16\). Since \(D < 0\), the equation has no real solutions.

a) \(c = 16\), and the solution is \(x = 4\). b) There are no real solutions.

- Set the height function equal to each target height. - Move all terms to one side before calculating the discriminant. - Use the sign of the discriminant to determine the number of times.

1. For part a), set \(h(t) = 21\): \(-16t^2 + 32t + 4 = 21\), so \(-16t^2 + 32t - 17 = 0\). 2. The discriminant is \(D = 32^2 - 4 \cdot (-16) \cdot (-17) = 1024 - 1088 = -64\). Since \(D < 0\), the ball does not reach \(21\,\text{ft}\). 3. For part b), set \(h(t) = 20\): \(-16t^2 + 32t + 4 = 20\), so \(-16t^2 + 32t - 16 = 0\). 4. The discriminant is \(D = 32^2 - 4 \cdot (-16) \cdot (-16) = 1024 - 1024 = 0\). Therefore, the ball is at \(20\,\text{ft}\) at exactly one time.

a) No; the discriminant is \(-64\). b) The ball is at \(20\,\text{ft}\) at exactly one time.

- Substitute the given value of \(c\) before solving parts a) and b). - For part c), write the discriminant in terms of \(c\). - Relate the sign of the discriminant to the number of real solutions.

1. When \(c = 50\), \(2x^2 - 50 = 0\), so \(x^2 = 25\). Thus, \(x = -5\) or \(x = 5\). 2. When \(c = 10\), \(2x^2 - 10 = 0\), so \(x^2 = 5\). Thus, \(x = -\sqrt{5}\) or \(x = \sqrt{5}\). 3. For \(2x^2 - c = 0\), the discriminant is \(D = 0^2 - 4 \cdot 2 \cdot (-c) = 8c\). 4. If \(c > 0\), then \(D > 0\), so there are two real solutions. If \(c = 0\), then \(D = 0\), so there is one real solution. If \(c < 0\), then \(D < 0\), so there are no real solutions.

a) \(\{-5, 5\}\) b) \(x = -\sqrt{5}\) or \(x = \sqrt{5}\) c) Two real solutions when \(c > 0\), one real solution when \(c = 0\), and no real solutions when \(c < 0\).

- Write the discriminant as an expression in \(c\). - Use \(D = 0\) for exactly one real solution. - Use \(D < 0\) to choose a value for part c).

1. The discriminant is \(D = 8^2 - 4 \cdot 2 \cdot c = 64 - 8c\). 2. When \(c = 5\), \(D = 64 - 8 \cdot 5 = 24\). Since \(D > 0\), the equation has two real solutions. 3. For exactly one real solution, set \(D = 0\): \(64 - 8c = 0\), so \(c = 8\). 4. For no real solutions, require \(D < 0\): \(64 - 8c < 0\), so \(c > 8\). One possible value is \(c = 10\).

a) \(D = 24\); two real solutions b) \(c = 8\) c) One possible value is \(c = 10\).

- Check whether equation (I) is a perfect-square trinomial. - For equation (II), identify \(a\), \(b\), and \(c\). - Recall how the sign of the discriminant determines the number of real solutions.

1. For part a, \(x^2 - 6x + 9 = (x - 3)^2\). Therefore, \((x - 3)^2 = 0\), so \(x = 3\). 2. For part b, use \(a = 1\), \(b = -6\), and \(c = 8\): \(x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 8}}{2} = \frac{6 \pm 2}{2}\). Thus, \(x = 4\) or \(x = 2\). 3. For equation (I), the discriminant is \((-6)^2 - 4 \cdot 1 \cdot 9 = 0\), so there is one repeated real solution. 4. For equation (II), the discriminant is \((-6)^2 - 4 \cdot 1 \cdot 8 = 4\), which is positive, so there are two distinct real solutions.

a) \(x = 3\) b) \(x \in \{2, 4\}\) c) Equation (I) has discriminant \(0\), while equation (II) has a positive discriminant.

- Compute \(b^2-4ac\) for each quadratic. - A positive discriminant gives two real zeros, a zero discriminant gives one distinct real zero, and a negative discriminant gives no real zeros. - After determining the number of real zeros, use an efficient method to find them when they exist.

1. For \(f\), the discriminant is \(b^2-4ac=10^2-4\cdot 1\cdot 21=16>0\), so there are two real zeros. Factoring gives \((x+3)(x+7)=0\), so the zeros are \(x=-3\) and \(x=-7\). 2. Write \(g(x)=-3x^2-6x-9\). Its discriminant is \(b^2-4ac=(-6)^2-4\cdot(-3)\cdot(-9)=-72<0\), so there are no real zeros. 3. For \(h\), the discriminant is \(b^2-4ac=(-4)^2-4\cdot 0.5\cdot 8=0\), so there is one distinct real zero. Multiplying \(0.5x^2-4x+8=0\) by \(2\) gives \(x^2-8x+16=(x-4)^2=0\), so the zero is \(x=4\) with multiplicity \(2\).

a) Two real zeros: \(x=-7\) and \(x=-3\) b) No real zeros c) One distinct real zero: \(x=4\), with multiplicity \(2\)

- Substitute \(k = 15\) before solving part a). - Rewrite the general equation in standard form. - Use the sign of its discriminant for parts b) and c).

1. When \(k = 15\), \(3x^2 - 12 = 15\), so \(3x^2 = 27\) and \(x^2 = 9\). Thus, \(x = -3\) or \(x = 3\). 2. Rewrite the general equation as \(3x^2 - (k + 12) = 0\). Its discriminant is \(D = 12(k + 12)\). 3. Exactly one real solution occurs when \(D = 0\), so \(k = -12\). Then \(x = 0\). 4. There are no real solutions when \(D < 0\), so \(k + 12 < 0\), or \(k < -12\).

a) \(\{-3, 3\}\) b) \(k = -12\) c) \(k < -12\)

- Substitute the given solution into the equation in part a). - Treat \(c = 0\) separately before dividing by \(c\). - For \(c \ne 0\), determine when \(\frac{18}{c}\) is negative.

1. Substitute \(x = 3\): \(c(3)^2 - 18 = 0\), so \(9c = 18\) and \(c = 2\). 2. When \(c = 2\), \(2x^2 - 18 = 0\), so \(x^2 = 9\). The second solution is \(x = -3\). 3. When \(c = 0\), the equation becomes \(-18 = 0\), which is false for every value of \(x\). 4. For \(c \ne 0\), the equation is \(x^2 = \frac{18}{c}\). It has no real solution when \(\frac{18}{c} < 0\), which occurs when \(c < 0\). 5. Including the separate case \(c = 0\), the equation has no real solution for \(c \le 0\).

a) \(c = 2\); the second solution is \(x = -3\). b) The equation becomes the false statement \(-18 = 0\). c) \(c \le 0\)

- Clear the fractions by multiplying by a common denominator. - Distribute the subtraction sign carefully in equation (2). - Use the sign of the isolated value of \(x^2\) to determine whether real solutions exist.

1. For equation (1), multiply by \(15\): \(3(x^2 + 1) + 5(2x^2 - 2) = 45\). 2. Simplify: \(13x^2 - 7 = 45\), so \(x^2 = 4\). Therefore, \(x = -2\) or \(x = 2\). 3. For equation (2), multiply by \(15\): \(3(x^2 + 1) - 5(2x^2 - 2) = 45\). 4. Simplify: \(-7x^2 + 13 = 45\), so \(x^2 = -\frac{32}{7}\). 5. A real number cannot have a negative square, so equation (2) has no real solution.

(1) \(x \in \{-2, 2\}\) (2) \(\varnothing\)

- Use the standard discriminant \(b^2 - 4ac\). - Set the discriminant equal to \(0\) for one real zero. - Use the opening direction to interpret a negative discriminant graphically.

1. The discriminant is \(D = (-6)^2 - 4 \cdot 1 \cdot k = 36 - 4k\). 2. For exactly one real zero, set \(D = 0\): \(36 - 4k = 0\), so \(k = 9\). 3. When \(k = 9\), \(f_k(x) = x^2 - 6x + 9 = (x - 3)^2\), so the vertex is \((3, 0)\). 4. When \(k = 10\), \(D = 36 - 4 \cdot 10 = -4\). Therefore, there are no real zeros. 5. The parabola opens upward, so having no real zeros means its entire graph lies above the x-axis.

a) \(D = 36 - 4k\) b) \(k = 9\); vertex \((3, 0)\) c) No real zeros; the graph lies entirely above the x-axis.

- Put the equation in standard form. - Express its discriminant in terms of \(d\). - Match each sign of the discriminant with the number of real solutions.

1. Write the equation in standard form: \(5x^2 + d - 20 = 0\). 2. Its discriminant is \(D = 0^2 - 4 \cdot 5 \cdot (d - 20) = 20(20 - d)\). 3. For exactly two real solutions, require \(D > 0\), so \(d < 20\). 4. For exactly one real solution, require \(D = 0\), so \(d = 20\). Then \(x = 0\). 5. For no real solutions, require \(D < 0\), so \(d > 20\).

a) \(d < 20\) b) \(d = 20\) c) \(d > 20\)

- Write the equation in standard quadratic form. - Calculate the discriminant in terms of \(a\) and \(b\). - Use the facts that \(a > 0\) and a real solution requires a nonnegative discriminant.

1. Expand and write the equation in standard form: \(ax^2 + a - b = 0\). 2. Since \(a > 0\), this is a quadratic equation with discriminant \(D = 0^2 - 4a(a - b) = 4a(b - a)\). 3. A real solution exists when \(D \ge 0\). Because \(a > 0\), this requires \(b - a \ge 0\), or \(b \ge a\). 4. Therefore, the claim is false. There is no real solution when \(b < a\). 5. For example, let \(a = 2\) and \(b = 1\). Then \(2(x^2 + 1) = 1\), which gives \(x^2 = -\frac{1}{2}\), so there is no real solution.

The claim is false. There is no real solution when \(b < a\). One counterexample is \(2(x^2 + 1) = 1\).

- Express the discriminant in terms of \(c\). - Set it equal to \(0\) in part a). - Determine the sign of the discriminant when \(c\) is greater than the value you found in part a).

1. The discriminant is \(D = (-10)^2 - 4 \cdot 1 \cdot c = 100 - 4c\). 2. For exactly one real solution, set \(D = 0\): \(100 - 4c = 0\), so \(c = 25\). 3. If \(c > 25\), then \(4c > 100\), so \(D = 100 - 4c < 0\). 4. Therefore, when \(c > 25\), the equation has no real solutions.

a) \(c = 25\) b) For \(c > 25\), the equation has no real solutions.

- Connect a perfect-square binomial with a repeated zero. - Set the discriminant equal to \(0\). - Remember both possibilities when taking the square root.

1. A quadratic expression is a perfect-square binomial when it has one repeated zero, so its discriminant is \(0\). 2. Here, \(D = (m + 3)^2 - 4 \cdot 4 \cdot 25 = (m + 3)^2 - 400\). 3. Set \(D = 0\): \((m + 3)^2 = 400\). 4. Therefore, \(m + 3 = 20\) or \(m + 3 = -20\). 5. Thus, \(m = 17\) or \(m = -23\). For these values, the expressions are \((2x + 5)^2\) and \((2x - 5)^2\), respectively.

\(m = 17\) or \(m = -23\)

- Set up the discriminant in terms of \(k\). - Simplify before solving \(D = 0\). - Verify that the leading coefficient is nonzero and factor the final expression.

1. A nonconstant quadratic is a perfect square of a linear expression when its discriminant is \(0\). 2. Calculate \(D = (2k)^2 - 4(k - 1)(k + 3)\). 3. Simplify: \(D = 4k^2 - 4(k^2 + 2k - 3) = -8k + 12\). 4. Set \(D = 0\): \(-8k + 12 = 0\), so \(k = \frac{3}{2}\). 5. The leading coefficient is then \(k - 1 = \frac{1}{2} \ne 0\), and \(f(x) = \frac{1}{2}x^2 + 3x + \frac{9}{2} = \frac{1}{2}(x + 3)^2\).

\(k = \frac{3}{2}\)

- Write the discriminant as an expression in \(k\). - Use \(D = 0\), \(D > 0\), and \(D < 0\) for the three cases. - Solve the resulting equations and inequalities.

1. The discriminant is \(D = k^2 - 4 \cdot 1 \cdot 25 = k^2 - 100\). 2. For exactly one real solution, \(D = 0\), so \(k = -10\) or \(k = 10\). 3. For two distinct real solutions, \(D > 0\), so \(k < -10\) or \(k > 10\). 4. For no real solutions, \(D < 0\), so \(-10 < k < 10\).

a) \(k = -10\) or \(k = 10\) b) \(k < -10\) or \(k > 10\) c) \(-10 < k < 10\)

- When does a quadratic equation have exactly one real solution? - How is a vertex on the \(x\)-axis related to the number of zeros? - What does the sign of the leading coefficient tell you about a parabola? - How does \((x - h)^2 + k\) expand into standard form?

1. The graph touches the \(x\)-axis at exactly one point when the quadratic equation has one real solution. Therefore, its discriminant must be zero: \(p^2 - 4q = 0\). Equivalently, \(q = \frac{p^2}{4}\). 2. A vertex at \((5, 0)\) gives \(f(x) = (x - 5)^2 = x^2 - 10x + 25\). Thus, \(p = -10\) and \(q = 25\). 3. The leading coefficient is \(1 > 0\), so the parabola opens upward. Its vertex is therefore a minimum, and the function has no maximum because its values increase without bound.

1) \(p^2 - 4q = 0\), equivalently \(q = \frac{p^2}{4}\) 2) \(p = -10\) and \(q = 25\) 3) The positive leading coefficient makes the parabola open upward, so its vertex is a minimum and there is no maximum.

- Use the discriminant to determine the number of real roots. - Compare \(t^2\) with \(64\). - Solve the resulting equality and inequalities carefully.

1. The discriminant of \(x^2+tx+16=0\) is \(D=t^2-4\cdot 1\cdot 16=t^2-64\). 2. There are two distinct real zeros when \(D>0\): \(t^2>64\), so \(t<-8\) or \(t>8\). 3. There is one repeated real zero when \(D=0\): \(t^2=64\), so \(t=-8\) or \(t=8\). 4. There are no real zeros when \(D<0\): \(t^2<64\), so \(-8<t<8\).

Two real zeros for \(t<-8\) or \(t>8\) One repeated real zero for \(t=-8\) or \(t=8\) No real zeros for \(-8<t<8\)

- Set the function equal to \(0\) and compute the discriminant. - The sign of the discriminant determines the number of x-intercepts. - Solve each condition for \(k\).

1. Set \(g_k(x)=0\): \(-x^2+4x+k=0\). The discriminant is \(D=4^2-4\cdot(-1)\cdot k=16+4k\). 2. There are two x-intercepts when \(D>0\): \(16+4k>0\), so \(k>-4\). 3. There is one x-intercept when \(D=0\): \(16+4k=0\), so \(k=-4\). 4. There are no x-intercepts when \(D<0\): \(16+4k<0\), so \(k<-4\).

Two x-intercepts for \(k>-4\) One x-intercept for \(k=-4\) No x-intercepts for \(k<-4\)

- Count each graph’s x-intercepts. - Connect two, one, or zero x-intercepts with the sign of the discriminant. - For part b), identify \(a\), \(b\), and \(c\) from the equation.

1. The graph of \(f\) crosses the x-axis twice, so \(f(x) = 0\) has two real solutions and \(D > 0\). 2. The graph of \(g\) stays above the x-axis, so \(g(x) = 0\) has no real solutions and \(D < 0\). 3. The graph of \(h\) touches the x-axis once, so \(h(x) = 0\) has one real solution and \(D = 0\). 4. For \(f(x) = -x^2 + 4x - 2\), \(D = 4^2 - 4 \cdot (-1) \cdot (-2) = 16 - 8 = 8\). 5. Since \(8 > 0\), the calculation agrees with the two x-intercepts shown in the graph.

a) \(f\): two real solutions and \(D > 0\) \(g\): no real solutions and \(D < 0\) \(h\): one real solution and \(D = 0\) b) \(D = 8\), which agrees with the graph.

- Set the two function expressions equal. - Put the resulting equation in standard quadratic form. - Determine when its discriminant is nonnegative.

1. Set the function values equal: \(x^2 + 4x + 7 = -x^2 + t\). 2. Write the resulting quadratic equation in standard form: \(2x^2 + 4x + (7 - t) = 0\). 3. Its discriminant is \(D = 4^2 - 4 \cdot 2 \cdot (7 - t) = 16 - 56 + 8t = 8t - 40\). 4. The graphs have at least one common point when \(D \ge 0\). 5. Solve \(8t - 40 \ge 0\): \(t \ge 5\).

The graphs touch or intersect for \(t \ge 5\).

- What must the vertex's \(x\)-coordinate be to lie on the \(y\)-axis? - How many real zeros does a quadratic have when its graph only touches the \(x\)-axis? - Which expression determines the number of real solutions of a quadratic equation?

1. The vertex of \(ax^2 + bx + c\) has \(x\)-coordinate \(-\frac{b}{2a}\). Here, \(x_v = -\frac{k}{4}\). For the vertex to lie on the \(y\)-axis, set \(-\frac{k}{4} = 0\), giving \(k = 0\). 2. A quadratic touches the \(x\)-axis at exactly one point when its discriminant is \(0\). Set \(k^2 - 4 \cdot 2 \cdot 8 = 0\). 3. Then \(k^2 - 64 = 0\), so \(k = 8\) or \(k = -8\). 4. In each case, the single point where the parabola touches the \(x\)-axis is the vertex, so the vertex lies on the \(x\)-axis.

a) \(k = 0\) b) \(k = 8\) or \(k = -8\); in either case, the vertex lies on the \(x\)-axis.

- Use the quadratic formula with the coefficients from the equation. - Multiply the conjugate radical expressions using the difference-of-squares pattern. - A quadratic has exactly one real solution when its discriminant equals zero.

1. For part a, use \(a = 1\), \(b = -1\), and \(c = -5\): \(x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-5)}}{2 \cdot 1} = \frac{1 \pm \sqrt{21}}{2}\). 2. For part b, multiply the solutions: \(\frac{1 + \sqrt{21}}{2} \cdot \frac{1 - \sqrt{21}}{2} = \frac{1 - 21}{4} = -5\). 3. For part c, write the equation as \(x^2 - x + c = 0\). It has exactly one real solution when the discriminant is zero: \((-1)^2 - 4 \cdot 1 \cdot c = 0\). 4. Solve \(1 - 4c = 0\), giving \(c = \frac{1}{4}\). Therefore, \(-5\) must be replaced by \(\frac{1}{4}\).

a) \(x = \frac{1 + \sqrt{21}}{2}\) or \(x = \frac{1 - \sqrt{21}}{2}\) b) The product is \(-5\). c) Replace \(-5\) with \(\frac{1}{4}\).

- Express the discriminant in terms of \(p\). - Match each required number of real solutions with a sign condition on the discriminant. - Solve the resulting quadratic inequalities carefully.

1. The discriminant is \(D = p^2 - 4 \cdot 1 \cdot 9 = p^2 - 36\). 2. For exactly one real solution, require \(D = 0\): \(p^2 - 36 = 0\), so \(p = -6\) or \(p = 6\). 3. For no real solutions, require \(D < 0\): \(p^2 < 36\), so \(-6 < p < 6\). 4. For two distinct real solutions, require \(D > 0\): \(p^2 > 36\), so \(p < -6\) or \(p > 6\).

a) \(p = -6\) or \(p = 6\) b) \(-6 < p < 6\) c) \(p < -6\) or \(p > 6\)

- Multiply by the least common denominator to clear the fractions. - Combine the \(x^2\)-terms carefully. - For part b, use the condition that a quadratic has exactly one real solution when its discriminant is zero.

1. For part a, multiply by \(12\): \(3(3x^2 + 5) - 4(2x^2 - 1) = 24\). 2. Simplify: \(9x^2 + 15 - 8x^2 + 4 = 24\), so \(x^2 = 5\). 3. Therefore, \(x = \pm\sqrt{5}\). 4. For part b, multiply the parameter equation by \(12\): \(x^2 + 19 = 12c\), so \(x^2 + 19 - 12c = 0\). 5. This quadratic has exactly one real solution when its discriminant is zero: \(0^2 - 4 \cdot 1 \cdot (19 - 12c) = 0\). 6. Solve \(19 - 12c = 0\): \(c = \frac{19}{12}\).

a) \(x \in \{-\sqrt{5}, \sqrt{5}\}\) b) \(c = \frac{19}{12}\)

- Build a quadratic whose roots have the given sum and product. - Use the standard discriminant \(b^2 - 4ac\). - A quadratic has one real root when its discriminant is zero.

1. Numbers with sum \(12\) and product \(k\) are the roots of \(z^2 - 12z + k = 0\). 2. For part a, set \(k = 32\): \(z^2 - 12z + 32 = 0\). Factor: \((z - 8)(z - 4) = 0\). Thus, the ordered pairs are \((8, 4)\) and \((4, 8)\). 3. For part b, the discriminant is \((-12)^2 - 4 \cdot 1 \cdot k = 144 - 4k\). 4. Exactly one real root occurs when \(144 - 4k = 0\), so \(k = 36\). 5. In that case, the repeated root is \(6\), and the only ordered-pair solution is \((6, 6)\).

a) \(\{(8, 4), (4, 8)\}\) b) \(k = 36\), giving the single ordered pair \((6, 6)\).

- Write the discriminant as a quadratic expression in \(k\). - Factor that expression. - Use a sign chart or the zeros of the factored expression to solve each inequality.

1. The discriminant is \(D = (k - 2)^2 - 4 \cdot 1 \cdot (k + 1)\). 2. Simplify: \(D = k^2 - 4k + 4 - 4k - 4 = k^2 - 8k = k(k - 8)\). 3. Exactly one real solution occurs when \(D = 0\), so \(k = 0\) or \(k = 8\). 4. No real solutions occur when \(D < 0\). The product \(k(k - 8)\) is negative between its zeros, so \(0 < k < 8\). 5. Two distinct real solutions occur when \(D > 0\), so \(k < 0\) or \(k > 8\).

No real solutions: \(0 < k < 8\) Exactly one real solution: \(k = 0\) or \(k = 8\) Two distinct real solutions: \(k < 0\) or \(k > 8\)

- Read each vertex directly from vertex form. - Eliminate the parameter to find the vertex locus. - For a fixed point \((x, y)\), treat the equation as a quadratic equation in \(a\). - Use the discriminant to decide when a real parameter value exists.

1. The equation is in vertex form, so the vertex of \(f_a\) is \((a, 2a)\). Eliminating \(a\) gives \(y=2x\). This is the locus of the vertices. 2. A fixed point \((x, y)\) lies on a graph in the family exactly when \(y=(x-a)^2+2a\) has a real solution for \(a\). Rearranging gives \(a^2+(2-2x)a+x^2-y=0\). Its discriminant is \(\Delta=(2-2x)^2-4(x^2-y)=4(1-2x+y)\). A real value of \(a\) exists exactly when \(\Delta\ge0\), which is equivalent to \(y\ge2x-1\). Therefore, no graph in the family passes through points satisfying \(y<2x-1\).

a) \(y=2x\) b) All points \((x, y)\) such that \(y<2x-1\).

- Use the positive leading coefficient and the absence of real zeros. - Write and factor the discriminant as an expression in \(k\). - Determine where the factored expression is negative.

1. The leading coefficient is positive, so the expression is positive for every real \(x\) exactly when the corresponding parabola has no real zeros. 2. Its discriminant is \(D = k^2 - 4 \cdot 1 \cdot (k + 3) = k^2 - 4k - 12\). 3. Require \(D < 0\): \(k^2 - 4k - 12 < 0\). 4. Factor: \((k - 6)(k + 2) < 0\). 5. This product is negative between its zeros, so \(-2 < k < 6\).

\(-2 < k < 6\)

- First identify when the coefficient of \(x^2\) is \(0\). - Treat that linear case separately. - For all other values, use the discriminant.

1. If \(a = 2\), the quadratic term disappears and the equation becomes \(4x + 1 = 0\). It has one solution, \(x = -\frac{1}{4}\). 2. If \(a \ne 2\), the equation is quadratic. Its discriminant is \(D = 4^2 - 4 \cdot (a - 2) \cdot 1 = 24 - 4a\). 3. When \(D = 0\), \(24 - 4a = 0\), so \(a = 6\). The quadratic has one repeated real solution. 4. When \(D > 0\), \(a < 6\). Excluding the linear case \(a = 2\), the equation has two distinct real solutions for \(a < 6\) and \(a \ne 2\). 5. When \(D < 0\), \(a > 6\), so there are no real solutions.

One real solution: \(a = 2\) or \(a = 6\) Two distinct real solutions: \(a < 6\) and \(a \ne 2\) No real solutions: \(a > 6\)

- Use the vertex and one point to determine \(f\). - Set \(f(x)\) equal to \(g_k(x)\). - Treat a possible linear case separately, then require a negative discriminant.

1. From the graph, \(f\) has vertex \((2, 2)\) and passes through \((0, 0)\). 2. Write \(f(x) = a(x - 2)^2 + 2\). Substituting \((0, 0)\) gives \(0 = 4a + 2\), so \(a = -\frac{1}{2}\). Thus, \(f(x) = -\frac{1}{2}(x - 2)^2 + 2 = -\frac{1}{2}x^2 + 2x\). 3. Set the functions equal: \(-\frac{1}{2}x^2 + 2x = kx^2 - 4\). 4. Rearrange: \(\left(k + \frac{1}{2}\right)x^2 - 2x - 4 = 0\). 5. If \(k = -\frac{1}{2}\), the equation is linear and has a real solution. Otherwise, its discriminant is \(D = (-2)^2 - 4 \cdot \left(k + \frac{1}{2}\right) \cdot (-4) = 16k + 12\). 6. No common points require \(D < 0\): \(16k + 12 < 0\), so \(k < -\frac{3}{4}\).

\(k < -\frac{3}{4}\)

- Determine the equation of the parabola from its vertex and another point. - Notice the common point through which every line in the family passes. - A substitution for \(x-2\) can simplify the intersection equation. - Use the discriminant condition for a quadratic equation with no real solutions.

1. The parabola has vertex \((2,1)\) and passes through \((0,2)\). Write \(f(x)=a(x-2)^2+1\). Substituting \((0,2)\) gives \(2=4a+1\), so \(a=0.25\). Thus, \(f(x)=0.25(x-2)^2+1\). 2. Set the functions equal: \(0.25(x-2)^2+1=m(x-2)\). 3. Let \(u=x-2\). The equation becomes \(0.25u^2-mu+1=0\). 4. Its discriminant is \(D=(-m)^2-4 \cdot 0.25 \cdot 1=m^2-1\). 5. There are no intersections when \(D<0\). Therefore, \(m^2-1<0\), which gives \(-1<m<1\).

\(-1<m<1\)

- Use the vertex and one additional point to determine \(p\). - Write the general vertex form for \(h\). - Set the functions equal and require the resulting equation to have no real solution.

1. From the graph, \(p\) has vertex \((2, 1)\) and passes through \((0, 3)\). 2. Write \(p(x) = a_p(x - 2)^2 + 1\). Substituting \((0, 3)\) gives \(3 = 4a_p + 1\), so \(a_p = \frac{1}{2}\). Thus, \(p(x) = \frac{1}{2}(x - 2)^2 + 1 = \frac{1}{2}x^2 - 2x + 3\). 3. A quadratic with vertex \((0, 5)\) has the form \(h(x) = ax^2 + 5\), where \(a \ne 0\). 4. Common points would satisfy \(ax^2 + 5 = \frac{1}{2}x^2 - 2x + 3\), or \(\left(a - \frac{1}{2}\right)x^2 + 2x + 2 = 0\). 5. If \(a = \frac{1}{2}\), this equation is linear and has a solution. If \(a \ne \frac{1}{2}\), its discriminant is \(D = 2^2 - 4 \cdot \left(a - \frac{1}{2}\right) \cdot 2 = 8 - 8a\). 6. The graphs have no common point when \(D < 0\), so \(8 - 8a < 0\), which gives \(a > 1\).

\(h(x) = ax^2 + 5\) for \(a > 1\)