Question

algebra: concepts and connections - plc
modeling and analyzing quadratic functions: quadratic functions
which function has a vertex on the y - axis?
$f(x)=(x + 1)(x - 2)$
$f(x)=(x - 2)(x + 2)$
$f(x)=x(x + 2)$
$f(x)=(x - 2)^2$

Explanation:

Step1: Recall vertex on y-axis rule

A quadratic function has its vertex on the y-axis if it is an even function, meaning $f(-x) = f(x)$, or when the linear term (coefficient of $x$) is 0 in standard form $f(x)=ax^2+bx+c$.

Step2: Expand each function

For $f(x)=(x+1)(x-2)$:

$$\begin{align*} f(x)&=x^2-2x+x-2\\ &=x^2 - x - 2 \end{align*}$$

Linear term coefficient: $-1
eq 0$

For $f(x)=(x-2)(x+2)$:

$$\begin{align*} f(x)&=x^2+2x-2x-4\\ &=x^2 - 4 \end{align*}$$

Linear term coefficient: $0$

For $f(x)=x(x+2)$:

$$\begin{align*} f(x)&=x^2+2x \end{align*}$$

Linear term coefficient: $2
eq 0$

For $f(x)=(x-2)^2$:

$$\begin{align*} f(x)&=x^2-4x+4 \end{align*}$$

Linear term coefficient: $-4
eq 0$

Step3: Identify valid function

Only $f(x)=(x-2)(x+2)$ has no linear term, so its vertex is on the y-axis.

Answer:

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