- Identify the leading term. - Use the parity of the degree and the sign of the leading coefficient.

1. The leading term is \(-0.5x^4\), which controls the end behavior. 2. The degree is even, so both ends point in the same direction. 3. The leading coefficient is negative, so both ends point downward. Therefore, \(g(x)\to-\infty\) as \(x\to\infty\) and as \(x\to-\infty\).

\(g(x)\to-\infty\) as \(x\to\infty\) and as \(x\to-\infty\).

- Trace the graph far to the right and far to the left. - For an even-degree polynomial, the sign of the leading coefficient determines whether both ends rise or both fall.

1. The right end rises without bound, so \(f(x)\to\infty\) as \(x\to\infty\). 2. The left end also rises without bound, so \(f(x)\to\infty\) as \(x\to-\infty\). 3. A fourth-degree polynomial has even degree. Since both ends rise, its leading coefficient is positive.

As \(x\to\infty\), \(f(x)\to\infty\). As \(x\to-\infty\), \(f(x)\to\infty\). The leading coefficient is positive.

- Read the direction of the graph at the far left and far right. - Opposite end directions indicate an odd degree.

1. The right end rises, so \(h(x) \to \infty\) as \(x \to \infty\). 2. The left end falls, so \(h(x) \to -\infty\) as \(x \to -\infty\). 3. Because the two ends move in opposite directions, the degree must be odd.

As \(x \to \infty\), \(h(x) \to \infty\). As \(x \to -\infty\), \(h(x) \to -\infty\). The degree is odd.

- Which term has the greatest exponent of \(x\)? - The leading coefficient includes the sign of that term.

1. The highest exponent is \(5\), so the leading term is \(-3x^5\). 2. The degree is \(5\), and the leading coefficient is \(-3\).

Leading term: \(-3x^5\); degree: \(5\); leading coefficient: \(-3\).

- What does the table tell you about the signs of the two coefficients? - For a positive or negative coefficient multiplying \(x^2\), what happens at large positive and negative inputs?

1. The table shows that \(f(2)=4\), so \(4a=4\) and \(a=1>0\). Its quadratic graph rises at both ends. 2. The table shows that \(g(2)=-4\), so \(4b=-4\) and \(b=-1<0\). Its quadratic graph falls at both ends.

Both ends of \(f\) rise; both ends of \(g\) fall.

- Look at the leftmost and rightmost parts rather than only the middle. - Compare the direction near each low point with the direction farther toward each edge.

1. On both the far left and far right of the displayed curve, the graph bends upward. 2. The local low points do not determine the direction of the ends. Both ends rise.

Both ends rise.

- Use the parity of the degree to determine symmetry. - Use the sign of the leading coefficient to determine whether the ends rise or fall. - Consider very large positive and negative inputs.

1. For \(f(x)=-5x^8\), the degree is even, so the graph is symmetric about the y-axis. The leading coefficient is negative, so both ends fall: \(f(x)\to-\infty\) as \(x\to\infty\) and as \(x\to-\infty\). 2. For \(g(x)=0.4x^3\), the degree is odd, so the graph is symmetric about the origin. The leading coefficient is positive, so the left end falls and the right end rises: \(g(x)\to\infty\) as \(x\to\infty\) and \(g(x)\to-\infty\) as \(x\to-\infty\).

a) Symmetric about the y-axis; \(f(x)\to-\infty\) as \(x\to\infty\) and as \(x\to-\infty\) b) Symmetric about the origin; \(g(x)\to\infty\) as \(x\to\infty\), and \(g(x)\to-\infty\) as \(x\to-\infty\)

- Start with an upward-opening quadratic. - Shift it vertically to change the number of x-intercepts. - Use the leading term to determine the direction of both ends.

1. One example with no real zeros is \(f(x)=x^2+1\). It is always at least \(1\), and both ends rise. 2. One example with exactly two real zeros is \(g(x)=x^2-4\). Its zeros are \(-2\) and \(2\), and both ends rise. 3. In general, the degree must be even so the two ends have the same direction, and the leading coefficient must be positive so both ends point upward.

a) One example is \(f(x)=x^2+1\). b) One example is \(g(x)=x^2-4\). c) \(n\) is even and \(a_n>0\).

- Focus on the leading term of each polynomial. - Compare the end behavior of even and odd powers. - Identify the highest power of \(x\) in each expression. - The sign of the leading coefficient determines whether matching ends rise or fall, not whether the two ends match.

1. A polynomial has the same behavior on both ends when its degree is even. 2. Function \(f\) has degree 4 and a negative leading coefficient, so both ends fall. 3. Function \(g\) has degree 5, so its ends have opposite behavior. 4. Function \(h\) has degree 6 and a positive leading coefficient, so both ends rise. 5. Function \(k\) has degree 7, so its ends have opposite behavior.

\(f\) and \(h\)

- Check whether every exponent of \(x\) is a nonnegative integer. - Identify the leading term of each polynomial. - Use the parity of the degree and the sign of the leading coefficient.

1. Function \(f\) is a polynomial of degree 6 with leading coefficient \(-4\). Because the degree is even and the leading coefficient is negative, \(f(x) \to -\infty\) as \(x \to \pm\infty\). 2. Expand \(g\): \(g(x) = \frac{1}{3}x^3 - 3x\). It is a polynomial of degree 3 with leading coefficient \(\frac{1}{3}\). Thus, \(g(x) \to \infty\) as \(x \to \infty\), and \(g(x) \to -\infty\) as \(x \to -\infty\). 3. Function \(h\) is not a polynomial because \(\sqrt{x} = x^{1/2}\) has a non-integer exponent.

a) Polynomial; degree \(6\); leading coefficient \(-4\); \(f(x) \to -\infty\) as \(x \to \pm\infty\) b) Polynomial; degree \(3\); leading coefficient \(\frac{1}{3}\); \(g(x) \to \infty\) as \(x \to \infty\), and \(g(x) \to -\infty\) as \(x \to -\infty\) c) Not a polynomial because of \(\sqrt{x}\)

- Identify the leading term of each polynomial. - Use the degree and leading coefficient. - Compare even and odd powers for negative inputs. - For a product, determine the leading term without fully expanding when possible.

1. For \(f\), the leading term is \(-x^4\). The degree is even and the leading coefficient is negative, so \(f(x) \to -\infty\) as \(x \to \pm\infty\). 2. Expanding \(g\) gives \(g(x) = \frac{1}{2}x^3 - 4.5x\). The degree is odd and the leading coefficient is positive, so \(g(x) \to \infty\) as \(x \to \infty\), and \(g(x) \to -\infty\) as \(x \to -\infty\). 3. Expanding \(h\) gives \(h(x) = -x^3 + 4x^2 - 4x\). The degree is odd and the leading coefficient is negative, so \(h(x) \to -\infty\) as \(x \to \infty\), and \(h(x) \to \infty\) as \(x \to -\infty\).

a) \(f(x) \to -\infty\) as \(x \to \pm\infty\) b) \(g(x) \to \infty\) as \(x \to \infty\), and \(g(x) \to -\infty\) as \(x \to -\infty\) c) \(h(x) \to -\infty\) as \(x \to \infty\), and \(h(x) \to \infty\) as \(x \to -\infty\)

- The terms may not be written in descending powers. - Use algebraic identities to identify the leading term efficiently. - Compare the effect of even and odd exponents on large negative inputs.

1. For \(p\), the leading term is \(0.2x^6\). Its degree is even and its leading coefficient is positive, so \(p(x) \to \infty\) as \(x \to \pm\infty\). 2. Simplify \(q\): \(q(x) = 1 - x^4\). Its degree is even and its leading coefficient is negative, so \(q(x) \to -\infty\) as \(x \to \pm\infty\). 3. For \(r\), the leading term is \(-x^5\). Its degree is odd and its leading coefficient is negative, so \(r(x) \to -\infty\) as \(x \to \infty\), and \(r(x) \to \infty\) as \(x \to -\infty\).

a) \(p(x) \to \infty\) as \(x \to \pm\infty\) b) \(q(x) \to -\infty\) as \(x \to \pm\infty\) c) \(r(x) \to -\infty\) as \(x \to \infty\), and \(r(x) \to \infty\) as \(x \to -\infty\)

- Identify the leading term of each polynomial. - Determine whether the leading term has even or odd degree. - Use the sign of the leading coefficient to decide whether each end rises or falls.

1. The leading term determines the end behavior of a polynomial function. 2. For \(f(x) = -0.1x^4 + 5x - 1\), the leading term is \(-0.1x^4\). Its degree is even and its coefficient is negative, so \(f(x) \to -\infty\) as \(x \to \infty\) and as \(x \to -\infty\). 3. For \(g(x) = \frac{2}{3}x^5 - 10x^2\), the leading term is \(\frac{2}{3}x^5\). Its degree is odd and its coefficient is positive, so \(g(x) \to \infty\) as \(x \to \infty\), and \(g(x) \to -\infty\) as \(x \to -\infty\).

a) \(f(x) \to -\infty\) as \(x \to \infty\) and as \(x \to -\infty\) b) \(g(x) \to \infty\) as \(x \to \infty\), and \(g(x) \to -\infty\) as \(x \to -\infty\)

- Identify the leading term of each polynomial. - Determine whether its degree is even or odd. - Determine whether its leading coefficient is positive or negative. - Use those two features to decide the direction of the left and right ends.

1. For \(a(x) = 5x^3 - x\), the leading term \(5x^3\) has odd degree and a positive coefficient. The right end rises and the left end falls, so the match is III. 2. For \(b(x) = -x^4 + 8x^2\), the leading term \(-x^4\) has even degree and a negative coefficient. Both ends fall, so the match is II. 3. For \(c(x) = -2x^5 + 1\), the leading term \(-2x^5\) has odd degree and a negative coefficient. The right end falls and the left end rises, so the match is IV. 4. For \(d(x) = 0.5x^2 + 3x\), the leading term \(0.5x^2\) has even degree and a positive coefficient. Both ends rise, so the match is I.

1. III 2. II 3. IV 4. I

- Identify the leading term of each polynomial. - Use whether the degree is even or odd to compare the two ends. - Use the sign of the leading coefficient to decide whether the right end rises or falls.

1. For \(a(x)\), the leading term is \(-x^4\). Its degree is even and its coefficient is negative, so both ends fall. This matches \(f_2(x) = -x^2\). 2. For \(b(x)\), the product of the leading terms is \(x^3\). Its degree is odd and its coefficient is positive, so the left end falls and the right end rises. This matches \(f_3(x) = x^3\). 3. For \(c(x)\), the leading term is \(-x^5\). Its degree is odd and its coefficient is negative, so the left end rises and the right end falls. This matches \(f_4(x) = -x^3\). 4. For \(d(x)\), the leading term is \(\frac{1}{2}x^6\). Its degree is even and its coefficient is positive, so both ends rise. This matches \(f_1(x) = x^2\).

a) \(f_2(x) = -x^2\) b) \(f_3(x) = x^3\) c) \(f_4(x) = -x^3\) d) \(f_1(x) = x^2\)

- Focus on the leading term of each polynomial. - To find the leading term of a product, multiply the leading terms of the factors. - Compare the degree and sign of the leading coefficient with the four reference functions.

1. The function \(p(x) = -x^3\) has odd degree and a negative leading coefficient. Therefore, \(p(x) \to -\infty\) as \(x \to \infty\), and \(p(x) \to \infty\) as \(x \to -\infty\). 2. For \(g(x)\), the leading terms of the factors are \(-x\) and \(x^2\), so the leading term of the product is \(-x^3\). Therefore, \(g\) has the same end behavior as \(p\). 3. The leading term of \(h(x)\) is \(-3x^4\). Its degree is even and its coefficient is negative, so both ends fall. This matches the end behavior of \(-x^2\).

a) \(p(x) \to -\infty\) as \(x \to \infty\), and \(p(x) \to \infty\) as \(x \to -\infty\) b) Yes. The leading term of \(g\) is \(-x^3\). c) \(-x^2\)

- Identify the leading term. - Determine whether the degree is even or odd. - Use the sign of the leading coefficient to decide which end rises.

1. The leading term is \(-2x^5\). 2. The degree is \(5\), which is odd, and the leading coefficient is \(-2\), which is negative. 3. Therefore, the right end falls and the left end rises: \(f(x) \to -\infty\) as \(x \to \infty\), and \(f(x) \to \infty\) as \(x \to -\infty\).

\(f(x) \to -\infty\) as \(x \to \infty\), and \(f(x) \to \infty\) as \(x \to -\infty\)

- Identify the leading term of each polynomial. - Compare whether the two degrees are even or odd. - Decide whether the lower-degree terms affect end behavior.

1. The leading term of \(g\) is \(-x^4\). Its degree is even and its coefficient is negative, so \(g(x) \to -\infty\) as \(x \to \infty\) and as \(x \to -\infty\). 2. The leading term of \(h\) is \(-0.5x^6\). Its degree is even and its coefficient is negative, so \(h(x) \to -\infty\) as \(x \to \infty\) and as \(x \to -\infty\). 3. Therefore, the two functions have the same end behavior.

Yes. For both functions, the degree is even and the leading coefficient is negative, so the function values approach \(-\infty\) at both ends.

- Identify the leading term of each polynomial. - Use the parity of the degree to decide whether the two ends point in the same or opposite directions. - Use the sign of the leading coefficient to decide whether the right end rises or falls. - Expand part d) enough to identify its leading term.

1. The leading term of \(f\) is \(-4x^3\). Its degree is odd and its leading coefficient is negative, so \(f(x)\to-\infty\) as \(x\to\infty\) and \(f(x)\to\infty\) as \(x\to-\infty\). 2. The leading term of \(g\) is \(\frac{1}{4}x^4\). Its degree is even and its leading coefficient is positive, so both ends rise. 3. The leading term of \(h\) is \(-x^5\). Its degree is odd and its leading coefficient is negative, so the left end rises and the right end falls. 4. Expanding gives \(k(x)=-x^4-x^2+2\), whose leading term is \(-x^4\). Its degree is even and its leading coefficient is negative, so both ends fall.

a) \(f(x)\to-\infty\) as \(x\to\infty\), and \(f(x)\to\infty\) as \(x\to-\infty\). b) \(g(x)\to\infty\) as \(x\to\infty\) and as \(x\to-\infty\). c) \(h(x)\to-\infty\) as \(x\to\infty\), and \(h(x)\to\infty\) as \(x\to-\infty\). d) \(k(x)\to-\infty\) as \(x\to\infty\) and as \(x\to-\infty\).

- Compare the signs of the outputs on the left and right sides of each panel. - What symmetry would you expect from an odd power? From an even power? - Compare the two remaining functions at an input greater than \(1\).

1. The graph of \(y=x^3\) has odd degree and is symmetric about the origin. Panel c) has this behavior, so function (1) matches panel c). 2. The graphs of \(y=x^4\) and \(y=x^6\) are both symmetric about the y-axis. 3. For \(x>1\), \(x^6>x^4\), while for \(0<x<1\), \(x^6<x^4\). Panel a) has the steeper growth outside \([-1,1]\) and lies closer to the x-axis inside that interval, so it represents \(y=x^6\). Panel b) represents \(y=x^4\).

(1) Panel c) (2) Panel b) (3) Panel a)

- Identify the degree and sign of the leading coefficient. - Compare the direction of the two ends of each graph. - Determine whether each function has y-axis symmetry or origin symmetry. - Use the y-intercept to confirm the match.

1. Function (I) has odd degree and a negative leading coefficient, so its left end rises and its right end falls. It is symmetric about the origin. This matches graph a. 2. Function (II) has even degree and a positive leading coefficient, so both ends rise. It is symmetric about the y-axis and has y-intercept \(1\). This matches graph b. 3. Function (III) has even degree and a negative leading coefficient, so both ends fall. It is symmetric about the y-axis and passes through the origin. This matches graph c.

(I) Graph a (II) Graph b (III) Graph c

- Read the direction of the graph at the far left and far right. - Describe how the function outputs change as the inputs get very large in each direction. - Compare the two end directions to the patterns for even and odd degrees. - Use the right-end direction to determine the sign of the leading coefficient.

1. On the left, the graph rises; on the right, it falls. Thus \(f(x)\to\infty\) as \(x\to-\infty\), and \(f(x)\to-\infty\) as \(x\to\infty\). 2. The two ends move in opposite directions, which is the end behavior of an odd-degree polynomial. 3. The right end falls as \(x\to\infty\), so the leading coefficient must be negative.

a) \(f(x)\to\infty\) as \(x\to-\infty\), and \(f(x)\to-\infty\) as \(x\to\infty\). b) 1. The ends move in opposite directions, so the degree is odd. 2. The right end falls, so the leading coefficient is negative.

- Compare y-axis symmetry with origin symmetry. - Use the negative leading coefficients to determine right-end behavior. - For graph \(f\), choose different positive even degrees and negative coefficients.

1. Graph \(f\) is symmetric about the y-axis, so its exponent is even. Graph \(g\) is symmetric about the origin, so its exponent is odd. 2. Because both coefficients are negative, both functions approach negative infinity as \(x\to\infty\). 3. Examples with even degree and negative leading coefficient include \(-x^2\), \(-0.5x^4\), and \(-2x^6\). Each is symmetric about the y-axis and falls on both ends.

1. Graph \(f\): even exponent; graph \(g\): odd exponent 2. Both functions approach \(-\infty\) as \(x\to\infty\). 3. Possible examples are \(-x^2\), \(-0.5x^4\), and \(-2x^6\).

- Decide whether the two ends move in the same or opposite directions. - Use that comparison to determine whether the degree is even or odd. - Use the direction of the right end to determine the sign of the leading coefficient.

1. For graph \(p\), the left end falls and the right end rises. Since the ends move in opposite directions, the degree is odd. Since the right end rises, the leading coefficient is positive. 2. For graph \(q\), both ends fall. Since the ends move in the same direction, the degree is even. Since both ends fall, the leading coefficient is negative.

\(p\): odd degree; positive leading coefficient \(q\): even degree; negative leading coefficient

- Compare the direction of each graph's left and right ends. - Use whether the degree is even or odd. - Use the sign of the leading coefficient to determine which end rises.

1. Function \(p_1\) has even degree and a positive leading coefficient, so both ends rise. This matches graph \(f\). 2. Function \(p_2\) has odd degree and a negative leading coefficient, so the left end rises and the right end falls. This matches graph \(g\). 3. Function \(p_3\) has odd degree and a positive leading coefficient, so the left end falls and the right end rises. This matches graph \(h\).

1. \(p_1\): graph \(f\) 2. \(p_2\): graph \(g\) 3. \(p_3\): graph \(h\)

- Look at where the graph heads on the far left and far right. - Decide whether the y-values increase or decrease without bound on each end. - Use the direction of the right end to determine the sign of the leading coefficient.

1. On the right, the graph falls without bound. Thus, \(x\to\infty\Rightarrow f(x)\to-\infty\). 2. On the left, the graph rises without bound. Thus, \(x\to-\infty\Rightarrow f(x)\to\infty\). 3. The opposite end behaviors indicate an odd degree. Because the right end falls, the leading coefficient is negative, so \(a_n<0\).

\(x\to\infty\Rightarrow f(x)\to-\infty\) \(x\to-\infty\Rightarrow f(x)\to\infty\) The leading coefficient is negative: \(a_n<0\).

- Compare whether the two ends move in the same or opposite directions. - Recall what matching end directions imply about the degree.

1. Both ends of the graph fall. 2. Because the two ends move in the same direction, the degree must be even. 3. The end behavior is \(p(x)\to-\infty\) as \(x\to\infty\) and as \(x\to-\infty\).

The degree is even. Also, \(p(x)\to-\infty\) as \(x\to\infty\) and as \(x\to-\infty\).

- Equal magnitudes lead to an equation involving both powers. - Because \(x\) is positive, dividing by a common positive power preserves the comparison. - Which power grows faster for large positive inputs?

1. For \(x>0\), equal magnitudes mean \(2x^5=100x^3\). Divide by \(2x^3>0\) to get \(x^2=50\), so \(x=5\sqrt{2}\). 2. For \(x>5\sqrt{2}\), \(2x^5>100x^3\). Thus, \(p(x)>0\); the fifth-degree term eventually dominates.

a) \(x=5\sqrt{2}\). b) Positive; \(2x^5\) is larger than \(100x^3\) for \(x>5\sqrt{2}\).

- Look at the signs of the outputs for positive and negative inputs. - Compare the table with the shapes on each side of the y-axis. - Do both ends of each candidate graph point in the same direction?

1. The outputs change sign with the input, and their magnitudes fit a cubic relationship: \(f(x)=x^3\) is a matching rule. 2. Graph a) is the positive cubic; its left end falls and its right end rises. Graph b) has both ends falling.

Graph a). The left end falls, and the right end rises.

- Substitute the given point into each proposed function form. - Even exponents are consistent with y-axis symmetry. - Compare the degrees to determine which magnitude grows faster for large positive \(x\).

1. For \(f\), \(-8=a(2^2)=4a\), so \(a=-2\). Thus, \(f(x)=-2x^2\). 2. For \(g\), \(-8=b(2^4)=16b\), so \(b=-0.5\). Thus, \(g(x)=-0.5x^4\). 3. The coefficients are \(-2\) and \(-0.5\). For \(x>2\), \(\frac{|g(x)|}{|f(x)|}=\frac{0.5x^4}{2x^2}=\frac{x^2}{4}>1\). This ratio increases as \(x\) increases, so \(|g(x)|\) grows faster.

1. \(f(x)=-2x^2\) 2. \(g(x)=-0.5x^4\) 3. The coefficients are \(-2\) and \(-0.5\). For \(x>2\), \(|g(x)|>|f(x)|\), and \(|g(x)|\) grows faster.

- Factor out the greatest common power of \(x\). - How does the degree affect a polynomial's growth for large positive inputs? - Test a value greater than the largest intersection point.

1. Set the functions equal: \(x^3=0.01x^5\). 2. Factor: \(0.01x^5-x^3=x^3(0.01x^2-1)=0\). 3. Thus, \(x=0\) or \(0.01x^2=1\). The latter gives \(x^2=100\), so \(x=-10\) or \(x=10\). 4. The function \(g\) has the higher-degree leading term, so it grows faster as \(x\to\infty\). In fact, after the last intersection, \(g(x)>f(x)\) for every \(x>10\).

a) \(x=-10, 0, 10\) b) \(g(x)=0.01x^5\) eventually remains greater. For all \(x>10\), \(g(x)>f(x)\).

- Identify the leading term of each polynomial. - Use whether the degree is even or odd. - Use the sign of the leading coefficient. - Compare with the end behavior of basic power functions such as \(x^2\), \(-x^2\), \(x^3\), and \(-x^3\).

1. The functions are \(g(x) = -2x^4 + 5x^2 - 1\) and \(h(x) = 4x^3 - 2x\). 2. Function \(g\) has even degree and a negative leading coefficient. Therefore, \(g(x) \to -\infty\) as \(x \to \infty\) and as \(x \to -\infty\). 3. Function \(h\) has odd degree and a positive leading coefficient. Therefore, \(h(x) \to \infty\) as \(x \to \infty\), and \(h(x) \to -\infty\) as \(x \to -\infty\).

a) \(g(x) = -2x^4 + 5x^2 - 1\); \(g(x) \to -\infty\) as \(x \to \pm\infty\) b) \(h(x) = 4x^3 - 2x\); \(h(x) \to \infty\) as \(x \to \infty\), and \(h(x) \to -\infty\) as \(x \to -\infty\)

- Consider the sign of a large negative input raised to an even or odd power. - Expand only enough to identify the leading term. - The first written term is not always the leading term. - Determine which degree and leading-coefficient combination makes both ends fall.

1. Both ends fall when the polynomial has even degree and a negative leading coefficient. 2. For \(p_1\), the leading term is \(-x^4\). Its degree is even and its leading coefficient is negative, so it meets the condition. 3. For \(p_2\), the leading term is \(-x^3\). Its degree is odd, so it does not meet the condition. 4. For \(p_3\), the leading term is \(-5x^6\). Its degree is even and its leading coefficient is negative, so it meets the condition. 5. For \(p_4\), expanding gives \(x^2 - x^4\), whose leading term is \(-x^4\). It meets the condition. 6. For \(p_5\), the leading term is \(-x^5\). Its degree is odd, so it does not meet the condition.

1), 3), and 4). The degree must be even, and the leading coefficient must be negative.

- Compare the end behavior of even and odd powers. - Determine how a negative leading coefficient reverses the parent power function. - Focus on the leading term of each expression.

1. Opposite end behavior requires an odd degree. Because the right end falls, the leading coefficient must be negative. 2. Function \(p\) has degree 3 and leading coefficient \(-0.5\), so it has the required behavior. 3. Function \(q\) has degree 4, so both ends have the same behavior. It does not qualify. 4. Expanding \(r\) gives \(r(x) = -x^3 + 2x^2\). It has degree 3 and leading coefficient \(-1\), so it qualifies.

1. The degree must be odd, and the leading coefficient must be negative. 2. \(p\) and \(r\)

- Determine the required sign of the leading coefficient for an odd-degree polynomial. - Use the y-intercept to determine the constant term. - Try a simple cubic containing only the cubic and constant terms. - Substitute the given point to determine the remaining coefficient.

1. A cubic with the stated end behavior must have a positive leading coefficient. 2. The y-intercept condition gives \(f(0) = -3\), so the constant term is \(-3\). 3. Choose the simple form \(f(x) = ax^3 - 3\), where \(a > 0\). 4. Use \(P(1, 0)\): \(0 = a(1)^3 - 3\), so \(a = 3\). 5. Therefore, one possible function is \(f(x) = 3x^3 - 3\).

One possible function is \(f(x) = 3x^3 - 3\).

- Y-axis symmetry restricts the powers that may appear. - Determine the required sign of the leading coefficient. - Use \(f(0)\) to find the constant term. - Substitute the zero and choose a convenient negative leading coefficient.

1. Y-axis symmetry allows the form \(f(x) = ax^4 + bx^2 + c\). 2. Both ends must fall, so the leading coefficient must satisfy \(a < 0\). 3. From \(f(0) = 2\), \(c = 2\). 4. From \(f(2) = 0\), \(16a + 4b + 2 = 0\). 5. Choose \(a = -1\). Then \(-16 + 4b + 2 = 0\), so \(b = 3.5\). 6. One possible function is \(f(x) = -x^4 + 3.5x^2 + 2\).

One possible function is \(f(x) = -x^4 + 3.5x^2 + 2\).

- Combine like powers of \(x\) first. - Check what happens to the fourth-degree term when \(k = -4\). - Both ends fall when the leading term has even degree and a negative coefficient. - Solve the resulting inequality for \(k\).

1. Combine like terms: \(h(x) = (4 + k)x^4 + x^3 - 2x^2\). 2. When \(k = -4\), the fourth-degree terms cancel, so \(h(x) = x^3 - 2x^2\). Its degree is odd and its leading coefficient is positive. Therefore, \(h(x) \to \infty\) as \(x \to \infty\), and \(h(x) \to -\infty\) as \(x \to -\infty\). 3. For both ends to fall, the fourth-degree leading coefficient must be negative. Thus, \(4 + k < 0\). 4. Solving gives \(k < -4\).

a) \(h(x) \to \infty\) as \(x \to \infty\), and \(h(x) \to -\infty\) as \(x \to -\infty\) b) \(k < -4\)

- Substitute the coordinates of the given point into the function. - Determine the degree from the factors. - Identify the leading term of the polynomial. - Compare the degree and sign of the leading coefficient with those of \(g\).

1. Use the point \(P(0, 12)\): \(f(0) = a(2)(-1)(-3) = 6a\). 2. Solve \(6a = 12\) to get \(a = 2\). 3. With \(a = 2\), the leading term of \(f\) is \(2x^3\). It has odd degree and a positive leading coefficient. 4. The leading term of \(g\) is \(5x^3\), which also has odd degree and a positive leading coefficient. Therefore, both functions approach \(\infty\) as \(x \to \infty\) and approach \(-\infty\) as \(x \to -\infty\). 5. Thus, \(a = 2\) satisfies both conditions.

Yes. The value is \(a = 2\).

- Substitute the given point to determine \(k\). - Determine the total degree from the factors. - Decide what sign the leading coefficient must have for both ends to fall. - Compare that required sign with the value of \(k\) from the point condition.

1. Use the point \(P(2, 24)\): \(f(2) = k(2^2 + 2)(2 - 4)^2 = k(6)(4) = 24k\). 2. Solve \(24k = 24\) to get \(k = 1\). 3. The polynomial has degree \(4\), and with \(k = 1\), its leading term is \(x^4\). 4. An even-degree polynomial with a positive leading coefficient approaches \(\infty\) at both ends, not \(-\infty\). 5. Therefore, no real value of \(k\) satisfies both conditions.

No. The point condition requires \(k = 1\), but then \(f(x) \to \infty\) as \(x \to \pm\infty\).

- Recall the end behavior of \(x^3\) and \(x^5\). - Imagine tracing the continuous graph from one end to the other. - Decide whether changing the sign of the leading coefficient affects the existence of an x-intercept.

1. When an odd-degree polynomial has a positive leading coefficient, \(g(x) \to \infty\) as \(x \to \infty\), and \(g(x) \to -\infty\) as \(x \to -\infty\). 2. More generally, every odd-degree polynomial has opposite end behavior, whether its leading coefficient is positive or negative. 3. Polynomial functions are continuous. Therefore, the graph must pass from negative function values to positive function values and cross the x-axis at least once. 4. Each x-intercept represents a real zero, so every odd-degree polynomial has at least one real zero.

a) \(g(x) \to \infty\) as \(x \to \infty\), and \(g(x) \to -\infty\) as \(x \to -\infty\) b) An odd-degree polynomial has opposite end behavior and is continuous, so its graph must cross the x-axis at least once.

- Use the volume formulas for a cube and a rectangular prism. - Equal volumes lead to an equation that can be factored. - Compare the degrees of the two volume functions for large \(x\).

1. The volume functions are \(V_C(x)=x^3\) and \(V_P(x)=4x^2\). 2. Set the volumes equal: \(x^3=4x^2\). 3. Factor: \(x^2(x-4)=0\). Since \(x>0\), \(x=4\,\text{m}\). 4. The common volume is \(4^3=64\,\text{m}^3\). 5. The cubic term \(x^3\) grows faster than the quadratic term \(4x^2\). In fact, \(x^3>4x^2\) for every \(x>4\).

a) \(V_C(x)=x^3\) and \(V_P(x)=4x^2\) b) \(x=4\,\text{m}\) c) \(64\,\text{m}^3\) d) The cubic function eventually exceeds the quadratic function; specifically, \(V_C(x)>V_P(x)\) for \(x>4\).

- Decide whether opposite end behavior occurs for an even or odd degree. - Use the right-end behavior to determine the sign of the leading coefficient. - The y-intercept is \(f(0)\). - Reflecting across the x-axis multiplies every function value by \(-1\).

1. The two ends move in opposite directions, so \(n\) must be odd. An even-degree polynomial has the same type of behavior on both ends. 2. The right end falls as \(x\to\infty\), so the leading coefficient must be negative: \(a_n<0\). 3. A cubic with a negative leading coefficient and constant term \(5\) is \(f(x)=-x^3+5\). Its constant term gives \(f(0)=5\). 4. Multiplying every output by \(-1\) reverses both end behaviors. Therefore, \(g(x)\to-\infty\) as \(x\to-\infty\), and \(g(x)\to\infty\) as \(x\to\infty\).

a) \(n\) must be odd. b) \(a_n<0\). c) One possible answer is \(f(x)=-x^3+5\). d) \(g(x)\to-\infty\) as \(x\to-\infty\), and \(g(x)\to\infty\) as \(x\to\infty\).

- Use the parity of the degree to identify symmetry. - Use the sign of the leading coefficient to determine the direction of the ends. - For the unused functions, determine whether outputs are restricted to one side of zero or cover all real numbers.

1. Graph I is symmetric about the y-axis and both ends fall. It must have even degree and a negative leading coefficient, so graph I represents \(f_2(x)=-0.5x^6\). 2. Graph II is symmetric about the origin and rises from left to right. It must have odd degree and a positive leading coefficient, so graph II represents \(f_3(x)=0.2x^5\). 3. For \(f_1(x)=1.5x^4\), the graph is symmetric about the y-axis and both ends rise. Its range is \([0, \infty)\). 4. For \(f_4(x)=-3x^3\), the graph is symmetric about the origin, rises on the left and falls on the right, and has range \(( -\infty, \infty)\).

a) Graph I: \(f_2(x)=-0.5x^6\); Graph II: \(f_3(x)=0.2x^5\) b) \(f_1\): even; both ends rise; range \([0, \infty)\). \(f_4\): odd; the left end rises and the right end falls; range \(( -\infty, \infty)\).

- Identify the leading term of each function. - Use whether the degree is even or odd to compare the two ends. - Use the sign of the leading coefficient to determine whether the right end rises or falls. - Compare these conclusions with the left and right edges of each graph.

1. The leading term of \(f\) is \(0.5x^3\). The degree is odd and the leading coefficient is positive, so the left end falls and the right end rises. This matches graph B. 2. The leading term of \(g\) is \(-0.25x^4\). The degree is even and the leading coefficient is negative, so both ends fall. This matches graph A. 3. The leading term of \(h\) is \(0.15x^4\). The degree is even and the leading coefficient is positive, so both ends rise. This matches graph C. 4. The leading term of \(k\) is \(-0.3x^3\). The degree is odd and the leading coefficient is negative, so the left end rises and the right end falls. None of the displayed graphs has this end behavior, so \(k\) is not used.

A: \(g(x)=-0.25x^4+2x^2\); B: \(f(x)=0.5x^3-2x\); C: \(h(x)=0.15x^4-1.5x^2+2\); not used: \(k(x)=-0.3x^3+1.5x\)

- Determine which function rules are even and which are odd. - Compare the direction of each graph's left and right ends. - Use the sign of the leading coefficient to choose between functions with the same symmetry.

1. Graph 1 has its left end falling and its right end rising, so it must have odd degree and a positive leading coefficient. It is also symmetric about the origin. Among the choices, only \(p(x) = x^3 - 3x\) has these properties. 2. Graph 2 has both ends falling, so it must have even degree and a negative leading coefficient. It is also symmetric about the y-axis. Among the choices, only \(r(x) = -x^4 + 2x^2\) has these properties.

a) Graph 1: \(p(x) = x^3 - 3x\); Graph 2: \(r(x) = -x^4 + 2x^2\) b) Graph 1 has odd-degree positive-leading-coefficient end behavior and origin symmetry. Graph 2 has even-degree negative-leading-coefficient end behavior and y-axis symmetry.

- Begin with the degree and sign of the leading coefficient. - Group the graphs by whether their ends rise or fall. - Check whether each function is even, odd, or neither. - Use the y-intercept to distinguish functions with the same end behavior.

1. Both \(g_1\) and \(g_2\) have odd degree and a positive leading coefficient, so their left ends fall and their right ends rise. They match panels a) and c). 2. The even-degree polynomial \(g_3\) has a positive leading coefficient, so both ends rise. It matches panel d). 3. The even-degree polynomial \(g_4\) has a negative leading coefficient, so both ends fall. It matches panel b). 4. Since \(g_1(0)=2\), \(g_1\) matches panel c). Since \(g_2(0)=0\), \(g_2\) matches panel a). The origin symmetry of \(g_2\) confirms this match.

a) \(g_1\): panel c) b) \(g_2\): panel a) c) \(g_3\): panel d) d) \(g_4\): panel b)

- Evaluate \(f(0)\) and compare it with each graph's y-intercept. - Use the leading term to determine the end behavior of \(f\). - Factor out the greatest common factor to solve \(f(x) = 0\). - Apply the zero-product property.

1. The y-intercept of \(f\) is \(f(0) = 0\), so its graph must pass through \((0, 0)\). Graph A has y-intercept \(2\), so graph A cannot represent \(f\). 2. The leading term of \(f\) is \(-x^4\). Its degree is even and its coefficient is negative, so \(f(x) \to -\infty\) as \(x \to \infty\) and as \(x \to -\infty\). Graph B rises at both ends, so graph B cannot represent \(f\). 3. Set the function equal to zero: \(-x^4 + 2x^2 = 0\). 4. Factor: \(-x^2(x^2 - 2) = 0\). 5. Therefore, \(x = 0\) or \(x^2 = 2\), which gives \(x = -\sqrt{2}\), \(x = 0\), and \(x = \sqrt{2}\).

a) Graph A has the wrong y-intercept, and graph B has the wrong end behavior. b) \(x = -\sqrt{2}, 0, \sqrt{2}\)

- Use the leading term to determine the end behavior of \(f\). - Evaluate \(f(0)\) and compare the y-intercepts. - Check \(f(-x)\) to determine symmetry. - Factor \(f\) and use the multiplicity of each zero.

1. The leading term of \(f\) is \(x^4\), so \(f(x) \to \infty\) as \(x \to \infty\) and as \(x \to -\infty\). Graph A falls at both ends, so it cannot represent \(f\). 2. Since \(f(0) = 0\), the graph of \(f\) passes through the origin. Graph B has y-intercept \(4\). It also has zeros at \(x = \pm 1\) and \(x = \pm 2\), which do not match the zeros of \(f\). Therefore, graph B cannot represent \(f\). 3. Because \(f(-x) = f(x)\), the graph is symmetric about the y-axis. 4. Factor the function: \(f(x) = x^2(x - 2)(x + 2)\). 5. The zero \(x = 0\) has multiplicity \(2\), so the graph touches the x-axis there. The zeros \(x = -2\) and \(x = 2\) have multiplicity \(1\), so the graph crosses the x-axis at those points.

a) Graph A has the wrong end behavior. Graph B has the wrong y-intercept and zeros. b) The graph is symmetric about the y-axis. It touches the x-axis at \(x = 0\) and crosses at \(x = -2\) and \(x = 2\).

- Begin with the degree and sign of the leading coefficient. - Compare the direction of the two ends of each graph. - Check for y-axis symmetry or origin symmetry. - Evaluate the y-intercept or identify a translated cubic's central point.

1. The function \(p\) has odd degree and a negative leading coefficient, so its left end rises and its right end falls. It is also symmetric about the origin. Therefore, \(p\) matches graph a. 2. The function \(q\) has even degree and a positive leading coefficient, so both ends rise. It is symmetric about the y-axis and has y-intercept \(q(0) = 1\). Therefore, \(q\) matches graph b. 3. The function \(r\) has odd degree and a positive leading coefficient, so its left end falls and its right end rises. It is a translation of \(y = x^3\) with central point \((1, -1)\). Therefore, \(r\) matches graph c.

\(p\): graph a; \(q\): graph b; \(r\): graph c

- Compare the left and right ends of each graph. - Use the degree and sign of each leading coefficient. - Check zeros or the y-intercept to confirm each match.

1. Graph a has its left end falling and its right end rising, so it must have odd degree and a positive leading coefficient. Only function 1 has this end behavior. 2. Function 1 factors as \(f(x) = 0.25x(x - 2)(x + 2)\), so its zeros are \(x = -2, 0, 2\), which agree with graph a. 3. Graph b falls at both ends, so it must have even degree and a negative leading coefficient. Only function 2 has this end behavior. 4. Function 2 factors as \(g(x) = -0.2x^2(x^2 - 6)\). The zero at \(x = 0\) has multiplicity \(2\), so the graph touches the x-axis at the origin, as graph b does.

Graph a: 1) \(f(x)=0.25x^3-x\); Graph b: 2) \(g(x)=-0.2x^4+1.2x^2\)

- Begin with the direction of the two ends of each graph. - Check for y-axis symmetry or origin symmetry. - Compare y-intercepts. - Use zeros and their multiplicities to confirm the matches.

1. Graph c is symmetric about the y-axis and falls at both ends. This matches \(p_3\), which has even degree and a negative leading coefficient. 2. Graph b has its left end rising and its right end falling. This matches \(p_2\), which has odd degree and a negative leading coefficient. Its double zero at \(x = 0\) and simple zero at \(x = 3\) confirm the match. 3. Graphs a and d both have odd-degree positive-leading-coefficient end behavior. 4. Graph a is symmetric about the origin and has zeros at \(x = 0\) and \(x = \pm\sqrt{3}\), so it matches \(p_1\). 5. Graph d has y-intercept \(-1\), a double zero at \(x = -1\), and a simple zero at \(x = 2\), so it matches \(p_4\).

1) \(p_1\): a 2) \(p_2\): b 3) \(p_3\): c 4) \(p_4\): d

- Compare whether the two ends move in the same or opposite directions. - Determine whether each graph has y-axis symmetry or origin symmetry. - Use the sign of the leading coefficient to decide which end behavior applies. - Identify the one unused rule by its unmatched end behavior.

1. Graph A is symmetric about the origin, with its left end falling and its right end rising. This matches function (1), which has odd degree and a positive leading coefficient. 2. Graph B is symmetric about the y-axis and falls at both ends. This matches function (4), which has even degree and a negative leading coefficient. 3. Graph C is symmetric about the origin, with its left end rising and its right end falling. This matches function (2), which has odd degree and a negative leading coefficient. 4. Function (3) is not used because its even degree and positive leading coefficient would make both ends rise.

a) A: (1); B: (4); C: (2); not used: (3) b) A and C have origin symmetry and opposite odd-degree end behavior. B has y-axis symmetry and both ends fall.

- Compare the left and right ends of the displayed graph with the leading term of \(f\). - Determine whether \(f\) is even, odd, or neither. - Factor \(f\) and examine the multiplicity of the zero at \(x = 0\). - Compare the graph and the function at a clearly readable integer x-value.

1. End behavior: The leading term of \(f\) is \(-0.5x^4\), so both ends of its graph fall. The displayed graph has its left end falling and its right end rising. 2. Symmetry: The function \(f\) is even because it contains only even powers, so its graph must be symmetric about the y-axis. The displayed graph is symmetric about the origin instead. 3. Behavior at the origin: Since \(f(x) = x^2(2 - 0.5x^2)\), \(x = 0\) is a zero of multiplicity \(2\). The graph of \(f\) must touch the x-axis at the origin, but the displayed graph crosses there. 4. Function value: \(f(2) = 0\), but the displayed graph passes through \((2, -2)\). 5. Any three of these contradictions are sufficient.

Possible reasons include: the end behavior is different; \(f\) has y-axis symmetry while the displayed graph has origin symmetry; \(f\) touches the x-axis at \(x = 0\) while the displayed graph crosses it; and \(f(2) = 0\) while the displayed graph has value \(-2\) at \(x = 2\). Any three are sufficient.

- Compare the degree and sign of each leading coefficient. - Use the y-intercept to distinguish graphs with similar shapes. - Check whether each function has y-axis symmetry or origin symmetry. - Look for zeros where a graph touches rather than crosses the x-axis.

1. The function \(f\) has odd degree and a positive leading coefficient, so its left end falls and its right end rises. It is also symmetric about the origin and passes through \((0, 0)\). This matches graph b. 2. The function \(g\) has even degree and a negative leading coefficient, so both ends fall. It has y-intercept \(g(0) = -1\). This matches graph a. 3. The function \(h\) has even degree and a positive leading coefficient, so both ends rise. It has y-intercept \(h(0) = 2\), and \(h(x) = \frac{1}{8}(x - 2)^2(x + 2)^2\), so it touches the x-axis at \(x = -2\) and \(x = 2\). This matches graph c.

a: \(g(x) = -\frac{1}{8}x^4 + x^2 - 1\); b: \(f(x) = \frac{1}{4}x^3 - x\); c: \(h(x) = \frac{1}{8}x^4 - x^2 + 2\)

- Identify the leading term of the polynomial. - Determine its sign for very large positive inputs. - Check a positive input outside the displayed window. - Could the graph turn after the visible interval ends?

1. The leading term is \(-0.01x^4\). For very large positive \(x\), this negative fourth-power term dominates, so \(f(x)\to-\infty\) as \(x\to\infty\). Emma’s claim is false. 2. Rewrite the function as \(f(x)=-0.01(x^2-25)^2+8.25\). This shows local maxima at \(x=-5\) and \(x=5\). 3. The graphing window ends exactly at those maxima. For \(x>5\), the graph turns downward, but that behavior is outside the displayed window. For example, \(f(10)=-48\).

Emma’s claim is false. \(f(x)\to-\infty\) as \(x\to\infty\). The graphing window ends at the local maximum \(x=5\), hiding the decrease for \(x>5\).

- Compare the y-intercepts by substituting \(x = 0\). - Determine whether each graph has y-axis symmetry or origin symmetry. - Compare the direction of the left and right ends.

1. Graph 1 is symmetric about the y-axis and rises at both ends, so it must have even degree and a positive leading coefficient. Its y-intercept is \(-2\). These features match \(f(x) = 0.5x^4 - 2\). 2. Graph 2 is symmetric about the origin, with its left end rising and its right end falling. It has zeros at \(x = -1, 0, 1\). These features match \(h(x) = -x^5 + x = -x(x - 1)(x + 1)(x^2 + 1)\).

Graph 1: \(f(x) = 0.5x^4 - 2\); Graph 2: \(h(x) = -x^5 + x\)

- Factor the polynomial to find its zeros. - Use the degree and leading coefficient to determine end behavior. - Compare these features with each displayed graph.

1. Factor the function: \(f(x) = -x(x^2 - x - 2) = -x(x - 2)(x + 1)\). The zeros are \(x = -1, 0, 2\). 2. The degree is \(3\) and the leading coefficient is negative. Therefore, \(f(x) \to \infty\) as \(x \to -\infty\), and \(f(x) \to -\infty\) as \(x \to \infty\). 3. Graph A has zeros at \(x = -2, 0, 1\), so its zeros do not match those of \(f\). 4. Graph B has the correct zeros, but its left end falls and its right end rises, which is the opposite of the end behavior of \(f\). 5. The graph comes from the upper left, crosses the x-axis at \(x = -1\), \(x = 0\), and \(x = 2\), and continues toward the lower right.

a) Graph A has the wrong zeros. Graph B has the correct zeros but the wrong end behavior. b) The graph crosses the x-axis at \((-1, 0)\), \((0, 0)\), and \((2, 0)\), with its left end rising and its right end falling.

- Use the exponents to determine symmetry. - Factor the polynomial and identify each zero's multiplicity. - Use the leading term to determine the end behavior.

1. The function contains only even powers, so its graph is symmetric about the y-axis. 2. Factor: \(f(x) = 0.5x^2(x - 2)(x + 2)\). The zero \(x = 0\) has multiplicity \(2\), and \(x = -2\) and \(x = 2\) are simple zeros. 3. The degree is \(4\) and the leading coefficient is positive, so both ends rise. 4. Graph A has four simple zeros at \(x = \pm 1\) and \(x = \pm 2\), and it does not have a zero at the origin. Therefore, it cannot represent \(f\). 5. Graph B has the correct zeros and touches at the origin, but both ends fall. Therefore, it cannot represent \(f\). 6. The graph is W-shaped and symmetric about the y-axis. It crosses at \(x = -2\) and \(x = 2\), touches at \(x = 0\), and rises at both ends.

a) Graph A has the wrong zeros. Graph B has the correct zeros but the wrong end behavior. b) The graph is symmetric about the y-axis, crosses at \((-2, 0)\) and \((2, 0)\), touches at \((0, 0)\), and rises at both ends.

- Compare the left and right ends of each graph. - Identify where each graph crosses or touches the x-axis. - Use a simple function value when needed to confirm a match. - Compare the degree with the overall graph shape.

1. Panel a) has odd-degree positive-leading-coefficient end behavior and zeros at \(x=-2,0,2\). It matches rule (1), since \(0.5x^3-2x=0.5x(x-2)(x+2)\). 2. Panel b) is a downward-opening parabola with vertex \((0,4)\) and zeros at \(x=\pm2\). It matches rule (2). 3. Panel c) has even-degree positive-leading-coefficient end behavior. It has a double zero at \(x=0\) and simple zeros at \(x=\pm2\). It matches rule (3), since \(x^4-4x^2=x^2(x-2)(x+2)\). 4. Rule (4) is not used because its left end rises and its right end falls.

a): rule (1) b): rule (2) c): rule (3) Not used: rule (4)

- Determine whether \(g\) is even or odd. - Evaluate \(g(0)\). - Use the degree and leading coefficient to determine the end behavior. - Compare each algebraic feature with the displayed graph.

1. Symmetry: The function \(g\) contains only even powers, so it is even and its graph must be symmetric about the y-axis. The displayed graph is not symmetric about the y-axis. 2. Y-intercept: \(g(0) = 3\), so the graph of \(g\) must cross the y-axis at \((0, 3)\). The displayed graph crosses at \((0, 0)\). 3. End behavior: The leading term of \(g\) is \(x^4\), so \(g(x) \to \infty\) as \(x \to -\infty\) and as \(x \to \infty\). The displayed graph falls to the left and rises to the right.

1. \(g\) is symmetric about the y-axis, but the displayed graph is not. 2. The y-intercept of \(g\) is \((0, 3)\), but the displayed graph has y-intercept \((0, 0)\). 3. Both ends of the graph of \(g\) rise, but the displayed graph falls to the left and rises to the right.

- Use the sign of the leading coefficient to determine end behavior. - Evaluate \(h(0)\). - Can you group the terms into an equivalent equation that looks easier to solve?

1. End behavior: The leading term is \(-0.5x^4\), so both ends of the graph of \(h\) must fall. The displayed graph rises at both ends. 2. Y-intercept: \(h(0) = -1\), but the displayed graph has y-intercept \(2\). 3. Zeros: Set \(-0.5x^4 + 2x^2 - 1 = 0\). Treating the equation as quadratic in \(x^2\) gives \(x^4-4x^2+2=0\), so \(x^2=2\pm\sqrt{2}\). Therefore, \(h\) has four real zeros: \(x=\pm\sqrt{2-\sqrt{2}}\) and \(x=\pm\sqrt{2+\sqrt{2}}\). The displayed graph has no x-intercepts.

The graph has the wrong end behavior, the wrong y-intercept, and the wrong number of real zeros.

- The equation for part a contains only even powers of \(x\). What simpler polynomial structure does that suggest? - For the end behavior, focus on the degree and sign of the leading coefficient. - Check whether the x-intercepts and the direction of both ends of the graph agree with your calculations.

1. Set \(f(x)=0\). Since the equation contains only \(x^4\) and \(x^2\), view it as a quadratic equation in \(x^2\): \(\frac{1}{2}x^4-3x^2+\frac{5}{2}=0\). 2. Multiply by \(2\): \(x^4-6x^2+5=0\). Factoring gives \((x^2-1)(x^2-5)=0\). 3. Thus \(x^2=1\) or \(x^2=5\), so the zeros are \(x=\pm1\) and \(x=\pm\sqrt{5}\). 4. The leading term is \(\frac{1}{2}x^4\). Its degree is even and its coefficient is positive, so \(f(x)\to\infty\) as \(x\to\infty\) and as \(x\to-\infty\). 5. The graph shows four x-intercepts near \(-\sqrt{5}\), \(-1\), \(1\), and \(\sqrt{5}\), and both ends rise.

a) The zeros are \(x=-\sqrt{5}\), \(x=-1\), \(x=1\), and \(x=\sqrt{5}\). b) \(f(x)\to\infty\) as \(x\to\pm\infty\).

- Determine the symmetry from the exponents. - Factor the function and identify the zeros and their multiplicities. - Use the squared form to determine whether the function can be negative. - Compare these properties with the displayed graph.

1. Symmetry: The function \(f\) contains only even powers, so its graph must be symmetric about the y-axis. The displayed graph is not symmetric about the y-axis. 2. Zeros and multiplicity: \(f(x) = (x^2 - 1)^2 = (x - 1)^2(x + 1)^2\). Therefore, the only zeros are \(x = -1\) and \(x = 1\), each with multiplicity \(2\), so the graph must touch the x-axis at those points. The displayed graph crosses the x-axis at four simple zeros: \(x = -2, -1, 1, 3\). 3. Sign of the function: Because \(f(x) = (x^2 - 1)^2\), \(f(x) \ge 0\) for every real \(x\). The displayed graph has portions below the x-axis.

The displayed graph has the wrong symmetry, the wrong zeros and multiplicities, and negative function values even though \(f(x) \ge 0\) for all real \(x\).

- Identify which functions contain only even powers and which contain only odd powers. - Use the degree to distinguish the parabola from the quartic graph. - Use the sign of the leading coefficient to distinguish the two cubic graphs.

1. The functions \(g_1\) and \(g_4\) contain only even powers, so both graphs are symmetric about the y-axis. The function \(g_1\) is quartic, which matches graph a, while \(g_4\) is an upward-opening parabola, which matches graph d. 2. The functions \(g_2\) and \(g_3\) contain only odd powers, so both graphs are symmetric about the origin. 3. The leading coefficient of \(g_2\) is positive, so its graph falls to the left and rises to the right. This matches graph c. 4. The leading coefficient of \(g_3\) is negative, so its graph rises to the left and falls to the right. This matches graph b.

\(a \to g_1\); \(b \to g_3\); \(c \to g_2\); \(d \to g_4\)

- Compare the end behavior of the three graphs. - Evaluate each function at \(x = 0\). - Check a visible zero to confirm the match. - Compare the relative steepness of graphs with the same end behavior.

1. Functions \(f\) and \(h\) have positive leading coefficients, so their left ends fall and their right ends rise. These match graphs \(a\) and \(c\). 2. Function \(g\) has a negative leading coefficient, so its left end rises and its right end falls. This matches graph \(b\). 3. Compare y-intercepts: \(f(0) = 2\), so \(f\) matches graph \(a\). Also, \(h(0) = 1\), so \(h\) matches graph \(c\). 4. The zero at \(x = 1\) for both \(f\) and \(h\) confirms the two matches.

\(a\): \(f(x) = x^3 - 3x + 2\); \(b\): \(g(x) = -x^3 + 3x + 2\); \(c\): \(h(x) = 0.5x^3 - 1.5x + 1\)

- Compare the sizes of the terms for small \(|x|\). - Identify the term that dominates for large \(|x|\). - Compare the scales of the two windows. - Use the leading term to determine end behavior.

1. In Window 1, where \(-2\le x\le2\), the term \(-2x^2+50\) dominates because \(\frac{1}{100}x^4\) is very small. The graph appears to be a downward-opening parabola. 2. In Window 2, the positive quartic term becomes dominant for larger \(|x|\), and the graph shows the W-shaped behavior of a positive-leading-coefficient quartic. 3. Window 1 can falsely suggest that the graph continues downward. In fact, the leading term \(\frac{1}{100}x^4\) determines the end behavior, so \(g(x)\to\infty\) as \(x\to\pm\infty\).

a) Window 1 looks like a downward-opening parabola; Window 2 shows a W-shaped quartic. b) The narrow window hides the eventual dominance of \(\frac{1}{100}x^4\). The actual end behavior is \(g(x)\to\infty\) as \(x\to\pm\infty\).

- Begin by calculating the two counts for the given grid sizes. - To compare the models for any positive \(n\), look at the sign of their difference. - Once you find the positive point where the difference changes sign, remember that \(n\) must be an integer.

1. At \(n=3\), \(R(3)=54+36=90\) and \(S(3)=90+12=102\). At \(n=4\), \(R(4)=128+64=192\) and \(S(4)=160+16=176\). 2. Subtract: \(R(n)-S(n)=2n(n^2-3n-2)\). For positive \(n\), the sign depends on \(n^2-3n-2\). 3. Its positive zero is \(\frac{3+\sqrt{17}}{2}\), between \(3\) and \(4\). Hence \(R(n)>S(n)\) exactly for integers \(n\ge4\). The cubic leading term of \(R\) eventually outweighs the quadratic leading term of \(S\).

a) At \(n=3\): \(R=90\), \(S=102\). At \(n=4\): \(R=192\), \(S=176\). b) All positive integers \(n\ge4\). The cubic leading term eventually dominates the quadratic terms.

- Identify which term is the leading term. - Both ends rise only for an even degree with a positive leading coefficient. - Consider separately the cases \(n < 6\), \(n = 6\), and \(n > 6\). - Watch for cancellation when \(n = 6\).

1. Both ends rise when the leading term has even degree and a positive coefficient. 2. If \(n < 6\), the leading term is \(-10x^6\), so both ends fall. No values of \(b\) work. 3. If \(n = 6\), then \(f(x) = (b - 10)x^6\). The coefficient must be positive, so \(b > 10\). 4. If \(n > 6\), the leading term is \(bx^n\), provided \(b \neq 0\). Both ends rise exactly when \(n\) is even and \(b > 0\). 5. Therefore, the conditions are \(n = 6\) with \(b > 10\), or an even integer \(n > 6\) with \(b > 0\).

\(n = 6\) and \(b > 10\), or \(n \in \{8, 10, 12, \ldots\}\) and \(b > 0\)

- Build a factored expression from the zeros and their multiplicities. - Use the given point to determine the leading constant. - End behavior depends on the degree and the sign of the leading coefficient. - Compare the left- and right-end behavior with that of \(-x^2\).

1. The zeros and multiplicities require the form \(f(x) = a(x + 1)^3(x - 2)\). 2. Use the point \((1, 16)\): \(16 = a(1 + 1)^3(1 - 2) = a(8)(-1)\). Thus, \(a = -2\). 3. The function \(f(x) = -2(x + 1)^3(x - 2)\) has degree \(3 + 1 = 4\) and leading coefficient \(-2\). 4. A polynomial of even degree with a negative leading coefficient approaches \(-\infty\) on both ends. Therefore, \(f(x) \to -\infty\) as \(x \to \pm\infty\), matching the end behavior of \(g(x) = -x^2\). 5. Such a function does exist.

Yes. One such function is \(f(x) = -2(x + 1)^3(x - 2)\). Its degree is even and its leading coefficient is negative, so \(f(x) \to -\infty\) as \(x \to \pm\infty\).

- Compare the two ends of the graph of an odd-degree polynomial. - Use the fact that polynomial functions are continuous. - Add the degrees of the factors to find the total degree. - Set each factor equal to zero and keep only real solutions. - Use each factor's exponent to determine multiplicity.

1. An odd-degree polynomial has opposite end behavior: one end approaches \(\infty\) and the other approaches \(-\infty\). 2. Polynomial functions are continuous. Therefore, as the graph moves from negative function values to positive function values, it must cross the x-axis at least once. Thus, every odd-degree polynomial has at least one real zero. 3. For \(g\), add the degrees of the factors: \(2 + 2 + 1 = 5\). Therefore, \(g\) has degree \(5\). 4. The factor \(x^2 + 4\) has no real zeros because \(x^2 + 4 > 0\) for every real \(x\). 5. The factor \((x - 5)^2\) gives the zero \(x = 5\) with multiplicity \(2\). 6. The factor \(x + 1\) gives the zero \(x = -1\) with multiplicity \(1\).

a) Because an odd-degree polynomial has opposite end behavior and is continuous, its graph must cross the x-axis at least once. b) Degree: \(5\); real zeros: \(x = 5\) with multiplicity \(2\), and \(x = -1\) with multiplicity \(1\)