SOLUTION: Find the range of the function \[h(x) = \frac{5x^2 + 20x + 33}{x^2 + 4x - 7}.\] Enter your answer in interval notation.

Question 1209685: Find the range of the function
\[h(x) = \frac{5x^2 + 20x + 33}{x^2 + 4x - 7}.\]
Enter your answer in interval notation.
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
To find the range of the function
\[h(x) = \frac{5x^2 + 20x + 33}{x^2 + 4x - 7},\]
we can rewrite it as
\[y = \frac{5x^2 + 20x + 33}{x^2 + 4x - 7}.\]
Then we have
\[y(x^2 + 4x - 7) = 5x^2 + 20x + 33\]
\[yx^2 + 4yx - 7y = 5x^2 + 20x + 33\]
\[yx^2 - 5x^2 + 4yx - 20x - 7y - 33 = 0\]
\[(y-5)x^2 + (4y-20)x - 7y - 33 = 0\]
\[(y-5)x^2 + 4(y-5)x - (7y+33) = 0\]
This is a quadratic equation in $x$. Since $x$ is real, the discriminant must be non-negative.
\[\Delta = [4(y-5)]^2 - 4(y-5)(-(7y+33)) \ge 0\]
\[16(y-5)^2 + 4(y-5)(7y+33) \ge 0\]
\[4(y-5)[4(y-5) + 7y+33] \ge 0\]
\[4(y-5)(4y - 20 + 7y + 33) \ge 0\]
\[4(y-5)(11y+13) \ge 0\]
\[(y-5)(11y+13) \ge 0\]
The roots are $y = 5$ and $y = -\frac{13}{11}$.
If $y < -\frac{13}{11}$ or $y > 5$, the inequality holds.
If $-\frac{13}{11} < y < 5$, the inequality does not hold.
If $y = 5$, we have
\[0x^2 + 0x - (35+33) = 0\]
\[-68 = 0\]
which is impossible.
If $y = -\frac{13}{11}$,
\[(-\frac{13}{11}-5)x^2 + 4(-\frac{13}{11}-5)x - (-7(-\frac{13}{11})+33) = 0\]
\[(-\frac{68}{11})x^2 + 4(-\frac{68}{11})x - (\frac{91}{11}+33) = 0\]
\[-68x^2 - 272x - 454 = 0\]
\[34x^2 + 136x + 227 = 0\]
$x = \frac{-136 \pm \sqrt{136^2 - 4(34)(227)}}{2(34)} = \frac{-136 \pm \sqrt{18496 - 30868}}{68}$
Since the discriminant is negative, there are no real solutions for $x$.
Therefore, the range is $(-\infty, -\frac{13}{11}) \cup (5, \infty)$.
Final Answer: The final answer is $\boxed{(-\infty, -13/11) \cup (5, \infty)}$

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