Question

complete the square to write $y = 3x^2 + 12x + 7$ in vertex form, $y = a(x - h)^2 + k$.
$y = 3(x^2 + 4x) + 7$
$y = 3(x^2 + 4x +4) + 7 - \square$
when the above expression is written in vertex form, $a$ is $\square$, $h$ is $\square$, and $k$ is $\square$.
done

Explanation:

Step1: Analyze the completion of square

When we add 4 inside the parentheses, we multiplied it by 3 (because of the factor 3 outside the parentheses). So the value we added is \(3\times4 = 12\), and to keep the equation balanced, we need to subtract 12. So the blank is 12.

Step2: Write the vertex form

Now, \(y = 3(x^{2}+4x + 4)+7-12\). We know that \(x^{2}+4x + 4=(x + 2)^{2}\), so \(y=3(x + 2)^{2}-5\). Comparing with \(y=a(x - h)^{2}+k\), we have \(a = 3\), \(h=- 2\) (since it's \(x-(-2)=x + 2\)), and \(k=-5\).

Answer:

The first blank: 12; \(a = 3\), \(h=-2\), \(k = - 5\)