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.

Respuesta :

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




for form [tex]f(x)=a(x-h)^2+k[/tex], the vertex is (h,k)

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

[tex]f(x)=a(x-1)^2-2[/tex]

find a by subsituting the given point

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

[tex]6=a(3-1)^2-2[/tex]

[tex]6=a(2)^2-2[/tex]

[tex]6=4a-2[/tex]

[tex]8=4a[/tex]

[tex]2=a[/tex]

[tex]f(x)=2(x-1)^2-2[/tex]

epxnad to find standard form

[tex]f(x)=2(x^2-2x+1)-2[/tex]

[tex]f(x)=2x^2-4x+2-2[/tex]

[tex]f(x)=2x^2-4x[/tex]


vertex form is [tex]f(x)=2(x-1)^2-2[/tex]

standard form is [tex]f(x)=2x^2-4x[/tex]

answer is A