Question: Review 1.4-1.6 APPC Name Key The graph of the polynomial function $f$ is shown above, where $-6 \leq x \leq 6$. Use the graph of $f$ to answer the following, or write “None” 1. $f$ has a local minimum at $x = 3$. 2. $f$ has a local maximum at $x = \underline{\hspace{1cm}}$. 3. The absolute maximum of $f$ is $\underline{\hspace{1cm}}$ at $x = \underline{\hspace{1cm}}$. 4. The absolute minimum of $f$ is $\underline{\hspace{1cm}}$ at $x = \underline{\hspace{1cm}}$. Graph shape: a polynomial curve on the top-right graph starts near $(-6, -3)$, dips to a local minimum around $(-4, -5)$, rises to a local maximum around $(0, -1)$, dips again to a local minimum around $(3, -4)$, then rises steeply to about $(6, 5)$; position_hint = top-right. 5. The graph of the polynomial function $k$ is shown above. Which of the following could be an expression for $k$? a. $-\frac{1}{20} (x + 4)^2 (x + 1) (x - 2)$ b. $\frac{1}{20} (x + 4)^2 (x + 1) (x - 2)$ c. $\frac{1}{20} (x - 4)^2 (x - 1) (x + 2)$ d. $\frac{1}{20} (x + 4) (x + 1) (x - 2)$ Graph shape: a polynomial curve on the lower-right graph has ends up on both sides, touches the x-axis near $x = -4$, crosses near $x = -1$ and $x = 2$, and has a local minimum around $(1, -2.5)$; position_hint = center. Find all real zeros and indicate the multiplicity of each zero. 6. $y = x^2(x + 2)^3(x - 6)^2$ 7. $k(x) = (x^2 - 9)(x^2 + 6x + 9)$

1. **Problem:** Identify local minima and maxima, absolute extrema of $f$ from the graph. 2. **Given:** Graph starts near $(-6, -3)$, local minimum near $(-4, -5)$, local maximum near $(0, -1)$, local minimum near $(3, -4)$, ends near $(6, 5)$. 3. **Step 1:** Local minima occur where the graph dips down and then rises. - At $x = -4$, local minimum near $-5$. - At $x = 3$, local minimum near $-4$ (given). 4. **Step 2:** Local maxima occur where the graph rises then dips. - At $x = 0$, local maximum near $-1$. 5. **Step 3:** Absolute maximum is the highest point on the graph. - At $x = 6$, $f(6) \approx 5$ is the highest point. 6. **Step 4:** Absolute minimum is the lowest point on the graph. - At $x = -4$, $f(-4) \approx -5$ is the lowest point. 7. **Answers for $f$:** - Local minimum at $x=3$ (given). - Local maximum at $x=0$. - Absolute maximum is $5$ at $x=6$. - Absolute minimum is $-5$ at $x=-4$. --- 8. **Problem:** Identify which polynomial expression matches the graph of $k$. 9. **Given:** Graph touches x-axis near $x=-4$ (implies even multiplicity zero), crosses at $x=-1$ and $x=2$ (odd multiplicity zeros), local minimum near $(1, -2.5)$. 10. **Step 1:** Check zeros and multiplicities in options: - a and b have $(x+4)^2$ (even multiplicity at $-4$), $(x+1)$ and $(x-2)$ linear factors. - c has $(x-4)^2$, $(x-1)$, $(x+2)$ zeros, which do not match graph zeros. - d has $(x+4)$ (multiplicity 1), so graph would cross at $-4$, but graph touches (even multiplicity). 11. **Step 2:** Check leading coefficient sign and end behavior. - Graph ends up on both sides, so leading coefficient positive. - So option b is positive coefficient with correct zeros and multiplicities. 12. **Answer:** Expression for $k$ is $\frac{1}{20} (x + 4)^2 (x + 1) (x - 2)$. --- 13. **Problem:** Find real zeros and multiplicities for $y = x^2 (x + 2)^3 (x - 6)^2$. 14. **Step 1:** Zeros are values of $x$ making factors zero. - $x=0$ with multiplicity 2. - $x=-2$ with multiplicity 3. - $x=6$ with multiplicity 2. 15. **Answer:** Zeros: $0$ (mult 2), $-2$ (mult 3), $6$ (mult 2). --- 16. **Problem:** Find real zeros and multiplicities for $k(x) = (x^2 - 9)(x^2 + 6x + 9)$. 17. **Step 1:** Factor each quadratic. - $x^2 - 9 = (x - 3)(x + 3)$ zeros at $3$ and $-3$, multiplicity 1 each. - $x^2 + 6x + 9 = (x + 3)^2$ zero at $-3$ multiplicity 2. 18. **Step 2:** Combine zeros and multiplicities. - Zero at $3$ multiplicity 1. - Zero at $-3$ multiplicity $1 + 2 = 3$ (since $(x+3)$ appears once and $(x+3)^2$ appears twice). 19. **Answer:** Zeros: $3$ (mult 1), $-3$ (mult 3).