What is the vertex form, f(x) = a(x − h)^2 + k, for a parabola that passes through the point (1, −7) and has (2, 3) as its vertex. What is the standard form of the equation?

A) Vertex form: f(x) = 10(x + 4)^2 + 5
Standard form: f(x) = −10x^2 + 40x + 165

B) Vertex form: f(x) = −6(x − 4)^2 + 3
Standard form: f(x) = −6x^2 + 7x − 24

C) Vertex form: f(x) = −10(x − 2)^2 + 3
Standard form: f(x) = −10x^2 + 40x − 37

D) Vertex form: f(x) = −3(x − 2)^2 + 4
Standard form: f(x) = −3x^2 + 12x +16

The vertex form of the equation is f(x)=-2(x-2)^2 - 4. The standard form of the equation is f(x)=-3x^2 + 12x - 16

ANSWER IS C

a is the correct

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!