What is the range of $f :R → R$, and $f(x) = x^2 + 6x − 8$ I have this discrete math question I have done completing the square but not sure how to continue. May I get some guide? Thanks!
What is the range of $f :R → R$, and $f(x) = x^2 + 6x − 8$
$f(x)=x^2+6x-8$
$f(x)=(x^2+6x+9)-8-9$
$f(x)=(x+3)^2-17$
 A: \begin{align}
f(x) & = x^2 + 6x - 8 \\
 & = (x^2 + 6x + 9) - 8 - 9 \\
 & = (x + 3)^2 - 17 \\
\end{align}
Thus, the range is $[-17, \infty)$, which follows immediately from the fact that $(x + 3)^2 \ge 0$ and that $f(x)$ is not bounded from above.
While this is the general approach to finding ranges of quadratic functions, consider this instead if you do not get the idea above:
Suppose that $f(x) = k$ where $x, k \in \Bbb R$, then the range of $f(x)$ is just the range of $k$.
\begin{align}
f(x) = k & \Leftrightarrow f(x) - k = 0 \\
 & \Leftrightarrow x^2 + 6x - (8 + k) = 0 \\
\end{align}
Since $x \in \Bbb R$, that is, $x^2 + 6x- (8 + k) = 0$ has real roots, which is equivalent to
\begin{align}
\Delta & = b^2 - 4ac \\
 & = 6^2 + 4 \cdot 1 \cdot (8 + k) \\
 & = 68 + 4k \\
 & \ge 0 \\
\end{align}
Thus,
$$f(x) = k \ge -17$$
and $[-17, \infty)$ is the range of $f(x)$.
A: From the above comments, we see that the range has to be $[-17, \infty)$ as the function is strictly increasing and unbounded above.
