Question

URGENT!! I have been trying to ask this question for a while but I keep getting fake answers: Find an equation in standard form for a parabolic function that goes through the points (1,-12) (3,16) and has a stretch factor of 2

Answer 1

Answer: y=a(x-h)^2+k

-12=a(1-h)^2+k

16=a(3-h)^2+k

subtract

a(3-h)^2-a(1-h)^2=28

a[(3-h)^2-(1-h)^2]=28

a=2

2[(3-h)^2-(1-h)^2]=28

(3-h)^2-(1-h)^2=14

9+h^2-6h-(1+h^2-2h)=14

9+h^2-6h-1-h^2+2h=14

-4h+8=14

-4h=6

h=-3/2

16=2(3+3/2)^2+k

16=2×81/4+k

k=16-81/2=-49/2

so y=2(x+3/2)²-49/2

Explanation:

Answer 2

Explanation:

y=a(x-h)^2+k

-12=a(1-h)^2+k

16=a(3-h)^2+k

subtract

a(3-h)^2-a(1-h)^2=28

a[(3-h)^2-(1-h)^2]=28

a=2

2[(3-h)^2-(1-h)^2]=28

(3-h)^2-(1-h)^2=14

9+h^2-6h-(1+h^2-2h)=14

9+h^2-6h-1-h^2+2h=14

-4h+8=14

-4h=6

h=-3/2

16=2(3+3/2)^2+k

16=2×81/4+k

k=16-81/2=-49/2

so y=2(x+3/2)²-49/2