Question

what is the range.f(x)=x^2+6x+8

Answer 1

To find the range of the function:

$f(x)=x^2+6x+8$

we first need to solve the equation:

$y=x^2+6x+8$

for x. Let's do this:

$\begin{gathered} y=x^2+6x+8 \\ y-8=x^2+6x \\ y-8+3^2=x^2+6x+3^2 \\ y-8+9=(x+3)^2 \\ y+1=(x+3)^2 \\ x+3=\pm\sqrt[]{y+1} \\ x=-3\pm\sqrt[]{y+1} \end{gathered}$

Now that we have an equation for x, we need to determine for which values of y the equation has real valued solutions. We notice that for this to be true we need that:

$\begin{gathered} y+1\ge0 \\ y\ge-1 \end{gathered}$

This values of y represent the range of the function. Therefore the range of function f is:

$\text{range}=\lbrack-1,\infty)$

This also can be obtanined by looking at the graph of the function:

(Remember that the range of the function is all the values the function takes in the y-axis)