1. **Problem Statement:** Given the graph of the function $y = f(x)$, answer the following questions based on the graph. 2. **(a) Evaluate $f(-2)$ and $f(6)$:** - From the graph, $f(-2) = 0$ (given). - $f(6)$ is not explicitly given, but from the graph near $x=6$, the value is approximately $3$ (given). 3. **(b) Find $f(2) - f(-6)$:** - From the graph, $f(2) = 3$ and $f(-6) = 8$. - Calculate: $$f(2) - f(-6) = 3 - 8 = -5$$ 4. **(c) State the zeros of the function:** - Zeros are where $f(x) = 0$. - From the graph, zeros occur at approximately $x=0$ and $x=4$. 5. **(d) Find the y-intercept:** - The y-intercept is the value of $f(0)$. - From the graph, $f(0) = 0$. 6. **(e) State the minimum and maximum values:** - Maximum value is at $(-7, 8)$. - Minimum value is at $(-4, -4)$ (note: graph shows minimum at $(-2, -5)$, but problem states $(-4, -4)$; assuming $(-2, -5)$ is correct from graph). 7. **(f) Coordinates of turning points:** - Maximum at $(-7, 8)$. - Minimum at $(-2, -5)$. 8. **(g) Interval where function is increasing:** - From $x = -2$ to $x = 5$, the function increases (from $-5$ to $5$). 9. **(h) Over interval $-5 < x < 4$, is the function positive or negative?** - From $x=-5$ to $x=0$, $f(x)$ is negative (below x-axis). - From $x=0$ to $x=4$, $f(x)$ is positive (above x-axis). - So, the function is negative for $-5 < x < 0$ and positive for $0 < x < 4$. **Final answers:** - $f(-2) = 0$ - $f(6) = 3$ - $f(2) - f(-6) = -5$ - Zeros at $x=0$ and $x=4$ - Y-intercept at $0$ - Maximum at $(-7, 8)$ - Minimum at $(-2, -5)$ - Increasing on interval $(-2, 5)$ - Function negative on $(-5, 0)$ and positive on $(0, 4)$