Answer:
Step-by-step explanation:
Hello, we know that if the equation is
[tex]y=a(x-h)^2+k[/tex]
Then the vertex is the the point (h,k)
Here, the vertex is the point (-2,-20) so we can write, a being a real number that we will have to find,
[tex]y=a(x-(-2))^2-20=a(x+2)^2-20[/tex]
On the other hand, we know that the y-intercept is (0,-12) so we can write
[tex]-20=a(0+2)^2-12=4a-12\\\\\text{We add 12 and we divide by 4.}\\\\4a = -20+12=-8\\\\a = \dfrac{-8}{4}=-2[/tex]
So the equation becomes.
[tex]\boxed{y=-2(x+2)^2-12}[/tex]
And we can give the standard form as below.
[tex]y=-2(x+2)^2-12=-2(x^2+4x+4)-12\\\\=-2x^2-8x-8-12 \ <=>\\\\\boxed{y=-2x^2-8x-20}[/tex]
Thank you.
Answer:
this is wrong it needs to be 2 positives
Step-by-step explanation:
