Create a quadratic equation that has an axis of symmetry of x = 1 and whose graph opens down. Then find the vertex, domain, range, and x-intercepts. Show all work to receive full credit.

Question

hero · Answer

I guess what I helped you with earlier was of no use.

anonymous · Answer

the quadratic equation needs to be in y = ax2 + bx + c form @Hero sorry

hero · Answer

Okay, well expand $(x-1)^2$ to get $x^2 -2x + 1$

hero · Answer

($x-1)^2 = x^2 - 2x + 1$

hero · Answer

$x^2 - 2x + 1$ is in the form $ax^2 + bx + c$. Do you agree?

anonymous · Answer

i agree

hero · Answer

Okay, so you're good then, right?

anonymous · Answer

i put it in a graphing calculator and it doesnt open down though :(

hero · Answer

I forgot that it had to be negative:

$-(x-1)^2 = -(x^2 - 2x +1)$ 

All you had to do was put a negative in front of it and you get the flip side.

hero · Answer

Or you can just input $-x^2 + 2x - 1$

anonymous · Answer

x has to equal one though..

hero · Answer

It will still equal 1. What me to prove it to you?

anonymous · Answer

no i believe u

hero · Answer

You have a calculator bro. Just graph and you will see that x still equals 1

anonymous · Answer

i mean squared

hero · Answer

I helped you with this earlier. If you don't know how to expand binomial squares, you need to brush up on your algebra.

anonymous · Answer

i would write it out like this right −1^2+2(1)−1=0 @Hero

hero · Answer

-(1)^2 + 2(1) - 1 = 0