Question

Use the function f(x) to answer the questions.

Answer 1

Answer:

See below

Step-by-step explanation:

Part A

Remember that x-intercepts are located where y=0:

$f(x)=-16x^2+60x+16$

$0=-16x^2+60x+16$

$0=x^2+60x-256$

$0=(x+64)(x-4)$

$0=(x-4)(-16x-4)$

$x=4$ and $x=-\frac{1}{4}$

Part B

Since the leading coefficient, -16, is negative, the parabola will open downward, making the vertex a maximum. We can determine the x-coordinate of the vertex using $x=-\frac{b}{2a}$ and plug it into the function to find the y-coordinate:

$x=-\frac{b}{2a}$

$x=-\frac{60}{2(-16)}$

$x=-\frac{60}{-32}$

$x=\frac{60}{32}$

$x=\frac{15}{8}$

Now we find the y-coordinate given the x-coordinate:

$f(x)=-16x^2+60x+16$

$f(\frac{15}{8})=-16(\frac{15}{8})^2+60(\frac{15}{8})+16$

$f(\frac{15}{8})=-16(\frac{225}{64})+\frac{900}{8}+16$

$f(\frac{15}{8})=\frac{-3600}{64}+\frac{225}{2}+16$

$f(\frac{15}{8})=\frac{-225}{4}+\frac{225}{2}+16$

$f(\frac{15}{8})=\frac{-225}{4}+\frac{450}{4}+\frac{64}{4}$

$f(\frac{15}{8})=\frac{289}{4}$

Therefore, the coordinates of the vertex are $(\frac{15}{8},\frac{289}{4})$.

Part C:

Notable points of a quadratic function include its x-intercepts, y-intercept, and vertex. Plotting these points on a graph help to visualize what the resulting graph of the function looks like. I've attached a graph with the function and the notable points for you to see.