Question

write the following quadratic function $6x^2 + 24x + y + 15 = 0$ in vertex form. $y = a(x - h)^2 + k$ $a = \square$ $h = \square$ $k = \square$ note: your answers should be integers.

Explanation:

Step1: Isolate y term

Start by moving the non - y terms to the other side of the equation. Given \(6x^{2}+24x + y+15 = 0\), we can rewrite it as \(y=-6x^{2}-24x - 15\).

Step2: Factor out the coefficient of \(x^{2}\) from the x - terms

For the quadratic part in x, which is \(-6x^{2}-24x\), we factor out the coefficient of \(x^{2}\) (which is - 6). So we have \(y=-6(x^{2}+4x)-15\).

Step3: Complete the square inside the parentheses

To complete the square for the expression \(x^{2}+4x\), we take half of the coefficient of x (half of 4 is 2), square it (\(2^{2}=4\)). We add and subtract this value inside the parentheses. But since there is a factor of - 6 outside the parentheses, we need to be careful with the arithmetic.

We rewrite \(x^{2}+4x\) as \(x^{2}+4x + 4-4=(x + 2)^{2}-4\).

Substitute this back into the equation for y:

\(y=-6((x + 2)^{2}-4)-15\)

Step4: Distribute and simplify

First, distribute the - 6:

\(y=-6(x + 2)^{2}+24-15\)

Then, simplify the constant terms: \(24-15 = 9\). So \(y=-6(x + 2)^{2}+9\).

Now, comparing with the vertex form \(y=a(x - h)^{2}+k\) (note that \(x+2=x-(-2)\)), we can see that:

• \(a=-6\)
• \(h=-2\)
• \(k = 9\)

Answer:

\(a=-6\), \(h=-2\), \(k = 9\)