Question

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

Answer 1

The vertex is a maximum because the leading coefficient is negative.

How to determine the x-intercept?

The function is given as:

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

Set the function to 0

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

Divide through by -4

4x^2 - 15x - 4 = 0

Expand

4x^2 - 16x + x - 4 = 0

Factorize

4x(x - 4) +1(x - 4) = 0

Factor out x - 4

(4x + 1)(x - 4) = 0

Solve for x

x = -1/4 and x = 4

Hence, the x-intercept of the graph of f(x) is -1/4 and 4

The vertex of the graph

The function is given as:

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

Differentiate

f'(x) = -32x + 60

Set to 0

-32x + 60 = 0

This gives

-32x = -60

Divide by -32

x = 1.875

Substitute x = 1.875 in f(x)=-16x^2 + 60x + 16

f(1.875) = -16 * 1.875^2 + 60* 1.875 + 16

Evaluate

f(1.875) = 72.25

Hence, the vertex of the graph of f(x) is (1.875, 72.25)

Also, the vertex is a maximum because the leading coefficient is negative.

Steps to graph f(x)

To graph f(x), we plot the x-intercepts and the vertex.

And then draw a curve through the points

See attachment for the graph

Read more about quadratic graphs at:

brainly.com/question/1214333

#SPJ1

Answer 2

Part A

$-16x^2 + 60x+16=0\\\\4x^2 - 15x-4=0\\\\(4x+1)(x-4)=0\\\\x=-\frac{1}{4}, 4$

Part B

The vertex of the graph will be a maximum because the leading coefficient is negative.

The x coordinate of the vertex is $-\frac{60}{2(-16)}=\frac{15}{8}$.

When x=15/8, $f\left(\frac{15}{8} \right)=\frac{289}{4}$.

So, the vertex has coordinates $\left(\frac{15}{8}, \frac{289}{4} \right)$

Part C

Plot the two x-intercepts and the vertex and then draw a curve in the shape of a parabola passing through them.

The graph is in the attached image.