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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 $f(x) = x^2 + 6x − 8$

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

$f(x)=(x^2+6x+9)-8-9$

$f(x)=(x+3)^2-17$

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    $\begingroup$ Just missing the last line, $$\implies f(x) \geq -17$$ $\endgroup$ – Nikunj Mar 29 '16 at 5:36
  • $\begingroup$ @Nikunj how do u get f(x)≥−17? $\endgroup$ – user292965 Mar 29 '16 at 5:39
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    $\begingroup$ That is because $(x+3)^2 \geq 0$ so the minimum value of $f(x)$ is $-17$ $\endgroup$ – Nikunj Mar 29 '16 at 5:40
  • $\begingroup$ @Nikunj, forgive me I really don't get why can I set it to 0. Is it a standard way? $\endgroup$ – user292965 Mar 29 '16 at 6:07
  • $\begingroup$ the value of $f(-3)$ is $-17$, then the function is strictly increasing. $\endgroup$ – aalo Mar 29 '16 at 6:25
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\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 $f(x)$.

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From the above comments, we see that the range has to be $[-17, \infty)$ as the function is strictly increasing and unbounded above.

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  • $\begingroup$ The function is strictly decreasing on the interval $(-\infty, -3)$, reaches its minimum of value of $-17$ at $x = -3$, and is strictly increasing on the interval $(3, \infty)$. It is unbounded above. $\endgroup$ – N. F. Taussig Mar 29 '16 at 18:22

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