Question

which function has a vertex on the y - axis?\\(\bigcirc\\ f(x) = (x - 2)^2\\)\\(\bigcirc\\ f(x) = x(x + 2)\\)\\(\bigcirc\\ f(x) = (x - 2)(x + 2)\\)\\(\bigcirc\\ f(x) = (x + 1)(x - 2)\\)

Explanation:

Step1: Recall vertex on y-axis rule

A quadratic function has its vertex on the y-axis if its axis of symmetry is $x=0$, meaning there is no $x$ term (linear term) when written in standard form $f(x)=ax^2+bx+c$ (so $b=0$).

Step2: Analyze Option 1

Expand $f(x)=(x-2)^2$:
$f(x)=x^2-4x+4$
Here, $b=-4
eq0$, so vertex not on y-axis.

Step3: Analyze Option 2

Expand $f(x)=x(x+2)$:
$f(x)=x^2+2x$
Here, $b=2
eq0$, so vertex not on y-axis.

Step4: Analyze Option 3

Expand $f(x)=(x-2)(x+2)$:
$f(x)=x^2-4$
Here, $b=0$, so vertex is on y-axis.

Step5: Analyze Option 4

Expand $f(x)=(x+1)(x-2)$:
$f(x)=x^2-x-2$
Here, $b=-1
eq0$, so vertex not on y-axis.

Answer:

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