Question

write the function $f(x)=x^2 + 8x - 48$ in vertex form.
$f(x)=a(x - h)^2 + k$
$a = \square$
$h = \square$
$k = \square$
note: your answers should be integers.

Explanation:

Step1: Identify the coefficient of \(x^2\)

The given function is \(f(x) = x^2 + 8x - 48\). In the vertex form \(f(x)=a(x - h)^2 + k\), the coefficient \(a\) is the same as the coefficient of \(x^2\) in the standard form. Here, the coefficient of \(x^2\) is \(1\), so \(a = 1\).

Step2: Complete the square for the \(x\)-terms

To complete the square for \(x^2 + 8x\), we take half of the coefficient of \(x\) (which is \(8\)), square it, and add and subtract it. Half of \(8\) is \(4\), and \(4^2 = 16\). So we rewrite the function as:

$$
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$$

In the vertex form \(f(x)=a(x - h)^2 + k\), comparing \((x + 4)^2 - 64\) with \(a(x - h)^2 + k\), we have \(h=-4\) (since \(x - h=x + 4\) implies \(h = - 4\)) and \(k=-64\).

Answer:

\(a = 1\), \(h=-4\), \(k = - 64\)