Question

what is the x-intercept of the graph of the function $f(x) = x^2 - 16x + 64$?
$\bigcirc$ $(-8, 0)$
$\bigcirc$ $(0, 8)$
$\bigcirc$ $(8, 0)$
$\bigcirc$ $(0, -8)$

Explanation:

Step1: Recall x-intercept definition

To find the x-intercept, set \( f(x) = 0 \), so we solve the equation \( x^{2}-16x + 64=0 \).

Step2: Factor the quadratic equation

The quadratic \( x^{2}-16x + 64 \) is a perfect square trinomial. We know that \( (a - b)^2=a^{2}-2ab + b^{2} \). Here, \( a = x \), \( 2ab=16x\Rightarrow b = 8 \), so \( x^{2}-16x + 64=(x - 8)^{2} \). So the equation becomes \( (x - 8)^{2}=0 \).

Step3: Solve for x

Taking the square root of both sides, we get \( x-8 = 0\Rightarrow x = 8 \). The x-intercept is the point where \( y = 0 \), so the x-intercept is \( (8,0) \).

Answer:

C. (8, 0)