Answer: Option C. f(x)=(x-2)(x+2)
A quadratic function f(x) has a vertex on the y-axis if the abcsissa of the vertex h is equal to zero
f(x)=ax^2+bx+c
h=-b/(2a)
A. f(x)=(x-2)^2
Using (a-b)^2=a^2-2ab+b^2
With a=x and b=2
f(x)=x^2-2(x)(2)+2^2
f(x)=x^2-4x+4
a=1, b=-4, c=4
h=-(-4)/[2(1)]=4/2→h=2 different of zero, then the function has not a vertex on the y-axis
B. f(x)=x(x+2)
Eliminating the parentheses using the distributive property in the multiplication:
f(x)=x x + 2x
f(x)=x^(1+1)+2x
f(x)=x^2+2x
a=1, b=2, c=0
h=-2/[2(1)]=-2/2→h=-1 different of zero, then the function has not a vertex on the y-axis
C. f(x)=(x-2)(x+2)
Eliminating the parentheses using (a-b)(a+b)=a^2-b^2; with a=x and b=2
f(x)=x^2-2^2
f(x)=x^2-4
a=1, b=0, c=-4
h=-0/[2(1)]=-0/2→h=0, then the function has a vertex on the y-axis
D. f(x)=(x+1)(x-2)
Eliminating the parentheses using (x+d)(x+e)=x^2+(d+e)x+d e; with d=1 and e=-2
f(x)=x^2+[1+(-2)]x+(1)(-2)
f(x)=x^2+(1-2)x-2
f(x)=x^2+(-1)x-2
f(x)=x^2-x-2
a=1, b=-1, c=-2
h=-(-1)/[2(1)]→h=1/2 different of zero, then the function has not a vertex on the y-axis