Question

I have a graph here. The vertex is
(4,-4), then the points (3,-3) and
(5,-3), then the points on the x axis are (2,0) and (6,0), then the points (1,5) and (5,5). This is a parabola that opens up.

I'm asked to write a function for f(x) in standard form.

I'm a little stuck. Any help you can provide is appreciated. Thank you.

Answer 1

You should be familiar with the vertex form of a parabola as
y = a(x-p)^2 + q , where (p,q) is the vertex,

so we have:
y = a(x-4)^2 - 4
using any of the other given points will give us the a
let's use the easiest, (2,0)

0 = a(2-4)^2 - 4
0 = 4a - 4
a= 1

y = (x-4)^2 - 4 is your equation.

A quick mental check shows that all your other points satisfy this equation.

Answer 2

(x-4)^2 = k (y+4) from vertex

put in some other point like (2,0)
(2-4)^2 = k (4)
4 = 4 k
k = 1

so I propose (x-4)^2 = y + 4
check with another given point like (5,5)
Hey (5,5) should be (7,5) !!!Typo?
Try (3,-3) and (5,-3)

Answer 3

Thank you!

Answer 4

You are welcome :)