Question

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Answer 1

We are given the following function:

$y=x^2+2x-8$

This is a quadratic equation since the greater exponent of the independent variable "x" is 2.

Every quadratic equation follows the general form:

$y=ax^2+bx+c$

To put it in the form:

$y=(x-h)^2+k$

We will use a technique called "completing the square". This consist in adding and subtracting the following term:

$(\frac{b}{2a})^2$

In the given equation these terms are:

$\begin{gathered} a=1 \\ b=2 \end{gathered}$

Therefore, the term we need to use is:

$(\frac{2}{2(1)})^2=1$

Therefore, we need to add and subtract 1 to the given function:

$y=x^2+2x-8+1-1$

Associating terms terms we get:

$\begin{gathered} y=x^2+2x+1+(-8-1) \\ y=(x^2+2x+1)-9 \end{gathered}$

Now we factor the expression inside the parenthesis using trinomial factoring. We take the square roots of the first and third term, we add them and we square them, like this:

$y=(x+1)^2-9$

And thus we have put the function in the desired form where:

$\begin{gathered} h=-1 \\ k=-9 \end{gathered}$