which function has a vertex on the y - axis?\n$f(x)=(x - 2)^2$\n$f(x)=x(x + 2)$\n$f(x)=(x - 2)(x +…

Explanation:

Step1: Recall vertex - form of quadratic function

The vertex - form of a quadratic function is (y = a(x - h)^2+k), where the vertex is ((h,k)). For a vertex on the (y) - axis, (h = 0). We can also expand the given functions and use the formula (x=-\frac{b}{2a}) for the (x) - coordinate of the vertex of the quadratic function (y = ax^{2}+bx + c).

Step2: Expand option A

Expand (f(x)=(x - 2)^2=x^{2}-4x + 4). Here, (a = 1), (b=-4). Using the formula (x=-\frac{b}{2a}), we get (x=-\frac{-4}{2\times1}=2), so the vertex is not on the (y) - axis.

Step3: Expand option B

Expand (f(x)=x(x + 2)=x^{2}+2x). Here, (a = 1), (b = 2). Using the formula (x=-\frac{b}{2a}), we get (x=-\frac{2}{2\times1}=-1), so the vertex is not on the (y) - axis.

Step4: Expand option C

Expand (f(x)=(x - 2)(x + 2)=x^{2}-4). Here, (a = 1), (b = 0), (c=-4). Using the formula (x=-\frac{b}{2a}), we get (x=-\frac{0}{2\times1}=0). Since the (x) - coordinate of the vertex is (0), the vertex lies on the (y) - axis.

Step5: Expand option D

Expand (f(x)=(x + 1)(x - 2)=x^{2}-x-2). Here, (a = 1), (b=-1). Using the formula (x=-\frac{b}{2a}), we get (x=-\frac{-1}{2\times1}=\frac{1}{2}), so the vertex is not on the (y) - axis.

Answer:

C. (f(x)=(x - 2)(x + 2))