How do I find the equation of a quadratic function with points at (-4,4) & vertex (-2,-4)? f(x)=?

Question

anonymous · Answer

We know the form

y=a(x-h)^2+k

(h,k) or (-2,-4) is the vertex, and since we have another point (x,y) or (-4,4) we can plug those in and solve for a.

4=a(-4+2)^2-4

anonymous · Answer

What do you get when you solve for a?

anonymous · Answer

4=4a-4
8=4a
2=a

anonymous · Answer

a=2

anonymous · Answer

right.

anonymous · Answer

Yes, now put that in.

y=2(x+2)^2-4

anonymous · Answer

If it wants standard form then multiply all of it out

anonymous · Answer

ok another example.
$$y=a(x-h)^2+k ... $$
Vertex (3,3) & point (5,11)
$$11=a(5-3)^2+3$$
$$11=a(2)^2+3$$
$$(8=4a)/4$$
$$a=2$$
So $$y=2(x-3)^2+3$$
Expanded
$$y=2x^2-36+3  ... becomes ...  y=2x^2-33$$

anonymous · Answer

ope ... 
$$y=2x^2-18+3$$ ... becomes ... $$y=2x^2-15$$

anonymous · Answer

i think you lost an x in there

anonymous · Answer

when expanding

anonymous · Answer

you should get 2x^2-12x+21

anonymous · Answer

... been a long time ... where did I go wrong?

anonymous · Answer

y=2(x−3)^2+3      first do (x-3)^2

y=2(x^2-6x+9)+3    distribute the 2

y=2x^2-12x+18+3     add the 3

y=2x^2-12x+21