Question

 Write the equation of the parabola that has the vertex at point (3,−12) and passes through the point (0,6).

Answer 1

Answer:

$y=2(x-3)^2-12$

Step-by-step explanation:

The vertex form of a parabola is

$y=a(x-h)^2+k$          ...(i)

where, a is a constant and (h,k) is vertex.

It is given that vertex of the parabola is (3,-12).

Substitute h=3 and k=-12 in equation (i).

$y=a(x-(3))^2+(-12)$

$y=a(x-3)^2-12$         ...(ii)

It is given that the parabola passes through the point (0,6). It means the equation of parabola must be true for (0,6).

Substitute x=0 and y=6 in equation (ii).

$6=a(0-3)^2-12$

$6+12=9a$

$18=9a$

$2=a$

Substitute a=2 in equation (ii).

$y=2(x-3)^2-12$

Therefore, the equation of parabola is $y=2(x-3)^2-12$.

Answer 2

Y=a (x-h)^2+k
y=a (x-3)^2+(-12)
plug in your point (0,6)
6=a (0-3)^2-12
6=a (9)-12
6=a (9) also seen as
6= 9a-12
now add 12 on both sides
18=9a
divide
2=a
y=2 (x-3)^2-12