Question

Find the equation f(x) = a(x - h)2 + k for a parabola containing point (3, 6) and having (1, -2) as a vertex. What is the standard form of the equation?
A) The vertex form is f(x) = 2(x - 1)2 − 2. The standard form is f(x) = 2x2−4x.
B) The vertex form is f(x) = 2(x + 1)2 + 2. The standard form is f(x) = −2x2−4x.
C) The vertex form is f(x) = −2(x - 7)2− 2. The standard form is f(x) = 2x2 + 4x.
D) The vertex form is f(x) = −2(x - 1)2 − 2. The standard form is f(x) = 2x2 +4 x.

Answer 1

the 1 is the h and the -2 is the k

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

f(x) = a(x-1)^2 -2

to find a  substitute x=3, y=6 in

6= a(3-1)^2 -2

6 =a *2^2 -2

add 2 to each side

8 = 4a

a =2

f(x) = 2(x-1)^2 -2  this is the vertex form

distribute to get the standard from

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

    2x^2 -4x+2-2

 f(x) =2x^2-4x   is the standard form

Choice A

Answer 2

for form $f(x)=a(x-h)^2+k$, the vertex is (h,k)

given that the vertex is (1,-2), h=1, k=-2

$f(x)=a(x-1)^2-2$

find a by subsituting the given point

(3,6), x=3 and f(x)=6

$6=a(3-1)^2-2$

$6=a(2)^2-2$

$6=4a-2$

$8=4a$

$2=a$

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

epxnad to find standard form

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

$f(x)=2x^2-4x+2-2$

$f(x)=2x^2-4x$

vertex form is $f(x)=2(x-1)^2-2$

standard form is $f(x)=2x^2-4x$

answer is A