Question

which function is equivalent to the function $f(x) = x^2 - 16x + 64$?
$\circ \\ f(x) = (x - 8)^2$
$\circ \\ f(x) = (x - 8)(x + 8)$
$\circ \\ f(x) = (x + 8)^2$
$\circ \\ f(x) = (x - 8)^2 + 64$

Explanation:

Step1: Recall the perfect square formula

The perfect square formula is \((a - b)^2 = a^2 - 2ab + b^2\).

Step2: Apply the formula to \(f(x)=x^2 - 16x + 64\)

In the function \(f(x)=x^2 - 16x + 64\), we can see that \(a = x\), and we need to find \(b\) such that \(-2ab=-16x\) and \(b^2 = 64\). From \(b^2 = 64\), we get \(b = 8\) (since \(-2\times x\times8=-16x\) matches the middle term). So using the perfect square formula, \(x^2 - 16x + 64=(x - 8)^2\).

Step3: Analyze other options

• For \(f(x)=(x - 8)(x + 8)\), expanding it gives \(x^2-64\), which is not equal to \(x^2 - 16x + 64\).
• For \(f(x)=(x + 8)^2\), expanding it gives \(x^2 + 16x + 64\), which is not equal to \(x^2 - 16x + 64\).
• For \(f(x)=(x - 8)^2+64\), expanding \((x - 8)^2\) gives \(x^2 - 16x + 64\), then adding 64 gives \(x^2 - 16x + 128\), which is not equal to \(x^2 - 16x + 64\).

Answer:

A. \(f(x)=(x - 8)^2\)