Question

find the axis of symmetry of the following quadratic function.
$f(x) = -(x + 2)(x - 2)$

Explanation:

Step1: Expand the quadratic function

First, we expand the given function \( f(x)=-(x + 2)(x - 2) \). Using the difference of squares formula \( (a+b)(a - b)=a^{2}-b^{2} \), we have \( (x + 2)(x - 2)=x^{2}-4 \). So the function becomes \( f(x)=-(x^{2}-4)=-x^{2}+4 \).

Step2: Recall the formula for the axis of symmetry of a quadratic function

For a quadratic function in the form \( f(x)=ax^{2}+bx + c \), the axis of symmetry is given by the formula \( x =-\frac{b}{2a} \).

Step3: Identify \( a \) and \( b \) from the expanded function

In the function \( f(x)=-x^{2}+4 \), we have \( a=-1 \) and \( b = 0 \) (since there is no \( x \) term, the coefficient of \( x \) is 0).

Step4: Calculate the axis of symmetry

Substitute \( a=-1 \) and \( b = 0 \) into the formula \( x=-\frac{b}{2a} \). We get \( x=-\frac{0}{2\times(-1)} = 0 \).

Alternatively, we can also find the axis of symmetry by finding the midpoint of the roots. For a quadratic function \( f(x)=a(x - r_1)(x - r_2) \), the roots are \( r_1 \) and \( r_2 \), and the axis of symmetry is the midpoint of the roots, given by \( x=\frac{r_1 + r_2}{2} \). For \( f(x)=-(x + 2)(x - 2) \), the roots are \( x=-2 \) and \( x = 2 \). The midpoint of -2 and 2 is \( \frac{-2+2}{2}=0 \), so the axis of symmetry is \( x = 0 \).

Answer:

The axis of symmetry is \( x = 0 \) (the y - axis).