Question

question 1-13
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 trinomial formula is \( (a - b)^2 = a^2 - 2ab + b^2 \).

Step2: Compare with the given function

For 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 \( b = 8 \) or \( b=-8 \).
If we take \( b = 8 \), then \( -2ab=-2\times x\times8=-16x \), which matches the middle term of the given quadratic function.
So, using the perfect square formula, \( x^{2}-16x + 64=(x - 8)^2 \).

Step3: Analyze other options

• For \( f(x)=(x - 8)(x + 8) \), using the difference of squares formula \( (a - b)(a + b)=a^{2}-b^{2} \), we get \( x^{2}-64 \), which is not equal to \( x^{2}-16x + 64 \).
• For \( f(x)=(x + 8)^2 \), using the perfect square formula \( (a + b)^2=a^{2}+2ab + b^{2} \), we get \( x^{2}+16x + 64 \), 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:

\( f(x)=(x - 8)^2 \) (the first option: \( f(x)=(x - 8)^2 \))