we know that
If the vertex is on the y-axis, then the x-coordinate of the vertex is equal to zero
we are going to verify the vertex of each one of the functions to determine the solution
Remember that
The equation in vertex form of a vertical parabola is equal to
[tex]y=a(x-h)^{2} +k[/tex]
where
(h,k) is the vertex
if [tex]a>0[/tex] -------> the parabola open upward (vertex is a minimum)
if [tex]a<0[/tex] -------> the parabola open downward (vertex is a maximun)
case A) [tex]f(x)=(x-2)^{2}[/tex]
This is a vertical parabola open upward
the vertex is the point [tex](2,0)[/tex]
therefore
The function [tex]f(x)=(x-2)^{2}[/tex] does not have a vertex on the y-axis
case B) [tex]f(x)=x(x+2)[/tex]
[tex]f(x)=x(x+2)=x^{2}+2x[/tex]
convert to vertex form
Complete the square. Remember to balance the equation by adding the same constants to each side.
[tex]f(x)+1=x^{2}+2x+1[/tex]
Rewrite as perfect squares
[tex]f(x)+1=(x+1)^{2}[/tex]
[tex]f(x)=(x+1)^{2}-1[/tex]
the vertex is the point [tex](-1,-1)[/tex]
therefore
The function [tex]f(x)=x(x+2)[/tex] does not have a vertex on the y-axis
case C) [tex]f(x)=(x-2)(x+2)[/tex]
[tex]f(x)=(x-2)(x+2)=x^{2}-2^{2}[/tex]
[tex]f(x)=x^{2}-4[/tex]
the vertex is the point [tex](0,-4)[/tex]
The x-coordinate of the vertex is equal to zero
therefore
The function [tex]f(x)=(x-2)(x+2)[/tex] has a vertex on the y-axis
case D) [tex]f(x)=(x+1)(x-2)[/tex]
[tex]f(x)=(x+1)(x-2)\\ \\f(x)= x^{2}-2x+x-2 \\ \\f(x)= x^{2} -x-2[/tex]
convert to vertex form
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex]f(x)+2= x^{2} -x[/tex]
Complete the square. Remember to balance the equation by adding the same constants to each side.
[tex]f(x)+2+0.25= x^{2} -x+0.25[/tex]
[tex]f(x)+2.25= x^{2} -x+0.25[/tex]
Rewrite as perfect squares
[tex]f(x)+2.25= (x-0.50)^{2}[/tex]
[tex]f(x)=(x-0.50)^{2}-2.25[/tex]
the vertex is the point [tex](0.5,-2.25)[/tex]
therefore
The function [tex]f(x)=(x+1)(x-2)[/tex] does not have a vertex on the y-axis
the answer is
[tex]f(x)=(x-2)(x+2)[/tex]
The vertex of the parabola is the point of intersection which divides the parabola into two equal parts.
The function which has the vertex on the y axis is,
[tex]f(x)=(x-2)(x+2)[/tex]
the vertex of the parabola is the point of intersection which divides the parabola into two equal parts.
The standard vertex form of the parabola is,
[tex]y=a(x-h)^2+k[/tex]
Here, [tex](h, k)[/tex] are the vertex of the parabola.
Given information-
The given function in the problem is,
[tex]f(x)=(x-2)^2[/tex]
Compare it with vertex form of parabola, we get the vertex as,
[tex](2,0)[/tex]
As the x coordinate is not zero. Thus this function does not has the vertex on the y axis.
[tex]f(x)=x(x+2)\\f(x)=x^2+2x\\f(x)=x^2+2x+1-1\\f(x)=(x+1)^2-1[/tex]
Compare it with vertex form of parabola, we get the vertex as,
[tex](-1,-1)[/tex]
As the x coordinate is not zero. Thus this function does not has the vertex on the y axis.
[tex]f(x)=(x-2)(x+2)\\f(x)=(x^2-2^2)\\f(x)=(x^2-4)[/tex]
Compare it with vertex form of parabola, we get the vertex as,
[tex](0,-4)[/tex]
As the x coordinate is zero. Thus this function has the vertex on the y axis
[tex]f(x)=(x+1)(x-2)\\f(x)=x^2-x-2\\f(x)=x^2-x-2+\dfrac{1}{4} -\dfrac{1}{4} \\f(x)=x^2-x+\dfrac{1}{4} -\dfrac{9}{4} \\f(x)=(x^2-\dfrac{1}{2}) -\dfrac{9}{4} \\[/tex]
Compare it with vertex form of parabola, we get the vertex as,
[tex]\dfrac{1}{2} ,\dfrac{-9}{4}[/tex]
As the y coordinate is not zero. Thus this function does not has the vertex on the y axis.
Hence the function which has the vertex on the y axis is,
[tex]f(x)=(x-2)(x+2)[/tex]
Learn more about the vertex of the parabola here
https://brainly.com/question/4563618
