Question

Sketch a graph representing f(x), indicating all intercepts and min or max points.

Answer 1

Explanations:

Given the function below;

$f(x)=4-\frac{x^2}{8}$

The x-intercept is the point where f(x) = 0 as shown:

$\begin{gathered} 0=4-\frac{x^2}{8} \\ 0=32-x^2 \\ x^2=32 \\ x=\pm\sqrt{32} \\ x=\pm5.66 \end{gathered}$

The x-intercept of the graph will be at (5.66, 0) and (-5.66, 0)

The y-intercept occurs at the. point where x = 0 to have:

$\begin{gathered} f(0)=4-\frac{0^2}{8} \\ f(0)=4 \end{gathered}$

The y-intercept is (0, 4)

The equivalent graphis as shown below

The graph shows that the function has a maxmum point at (0, 4) with no minimum point.