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Alex777 [14]
3 years ago
10

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

Mathematics
1 answer:
PIT_PIT [208]3 years ago
3 0

Answer:

See below

Step-by-step explanation:

<u>Part A</u>

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

<u />

<u />f(x)=-16x^2+60x+16<u />

<u />0=-16x^2+60x+16<u />

<u />0=x^2+60x-256<u />

<u />0=(x+64)(x-4)<u />

<u />0=(x-4)(-16x-4)<u />

<u />x=4 and x=-\frac{1}{4}

<u>Part B</u>

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}).

<u>Part C:</u>

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.

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