Question

Helpppp!
An expression is shown below:

f(x) = −16x^2 + 60x + 16


Is the vertex of the graph of f(x) going to be a maximum or minimum? What are the coordinates of the vertex? Justify your answers and show your work

&

What are the steps you would use to graph f(x)? Justify that you can use the answers obtained in Part A and Part B to draw the graph.

Answer 1

Are you allowed to use derivatives?

Answer 2

Yes, this is just a study guide. I would like an explanaion on how to find vertex and the other stufff. Ive got part a figured out. x1= -1/4 x2=4

Answer 3

ok. the vertex of the graph is (p,q), the x-coordinate "p" can be found by solving the solution: f'(x) = 0.
The fact that the coefficient of the term x^2 is negative (-16) tells you that the parabola looks like

Answer 4

solving the equation* when you find \(p\), the value \(q\) is easy to find because \((p,q)\) is on the graph: \(q=f(p)\).

Answer 5

the derivative of \(f\): \(f'(x) = (-16 x^2 + 60x+16)' = -16\cdot (2x) + 60\cdot 1\).

Answer 6

The other method (without derivatives) tells you that \(-\frac{b}{2a}\) is the x-coordinate of the vertex. The y-coordinate (q) is computed the same way as above: q = f(-b/(2a)).

Answer 7

How would you graph the function ?

Answer 8

I would first plot the x-intercepts and the maximum and then draw prabola through the points..

Right ?

Answer 9

yes

Answer 10

That would be it for The last question(:

Answer 11

yes