Range of $ y = \frac{x^3+3x^2+7x-11}{x^2+5x-6} $?

Question

How do I find the range of $ y = \frac{x^3+3x^2+7x-11}{x^2+5x-6} $? Would someone please help me understand how to do this by the method of forming a quadratic equation in x? A video lesson I was watching first factorises the numerator and denominator and finds that (x-1) is a common factor. It then forms a quadratic equation in x and finds values of y for discriminant greater than or equal to zero, getting a range R. It removes the (x-1) factor from numerator and denominator. then substitutes x=1 to get a y value 16/7. I don't get why this step works. It then equates 16/7 to the original expression to find x values,one of which is one and one of which is -19/7. It concludes that range is finally R. There is also a verification step which doesn't apply to this question but to other similar questions in which we try to get coefficient of $x^2 = 0$ and then check that y value. I don't get the purpose of any of this. I tried going through this question but it seems to deal with complex numbers and doesn't answer my doubts. What is the range of $f(x) = \frac{x^2-5x+6}{x-3}$?

Answer 1

$$ \frac{x^2 + 4x+11}{x+6} = k \; \; , \; \; x \neq 1 $$

$$ x^2 + 4x+11 = kx+6k \; \; , \; \; x \neq 1 $$ $$ x^2 + (4-k)x+(11-6k) = 0 \; \; , \; \; x \neq 1 $$ This has (real) solutions $x$ when the discriminant is at least zero, $$ (4-k)^2 - 4(11-6k) \geq 0, $$ $$ k^2 - 8k+ 16 +24k - 44 \geq 0, $$ $$ k^2 + 16k -28 \geq 0, $$ $$ k^2 + 16k +64 \geq 92, $$ $$ (k+8)^2 \geq 92. $$ Either $$ k+8 \geq \sqrt{92} $$ or $$ k+8 \leq -\sqrt{92} $$ This last pair is the same information you would get from the quadratic formula applied to $k^2 + 16 k - 28.$

I don't recall what happened with the $x \neq -1,$ you should draw some graphs

Note that $$ y = \frac{x^2 + 4x+11}{x+6} $$ describes a hyperbola, vertical asymptote at $x=-6,$ the other asymptote slanted (line $y=x-2$).

Answer 2

We can write $$\frac{x^3+3x^2+7x-11}{x^2+5x-6}=\frac{x^2+4x+11}{x+6}=x-2+\frac{23}{x+6},\quad x\ne 1. $$ If $x+6>0$, then by AM-GM inequality, we have $$ x-2+\frac{23}{x+6}=(x+6)+\frac{23}{x+6}-8\ge -8+2\sqrt{23}. $$ (Equality holds for $x+6=\sqrt{23}$.) We can observe that the value of the function taken by $x+6=7$ is also attained by $x+6=\frac{23}{7}$. Since it holds that $$\lim_{x\to\infty}\left(x-2+\frac{23}{x+6}\right)=\lim_{x\to -6^+}\left(x-2+\frac{23}{x+6}\right)=\infty,$$ the intermediate value theorem implies that the range of the given function on $\{x+6>0, x\ne 1\}$ is $$[-8+2\sqrt{23},\infty).$$

On the other hand, if $x+6<0$, then $$ (-x-6)+\left(-\frac{23}{x+6}\right)\ge 2\sqrt{23} $$ and $$ (x+6)+\frac{23}{x+6}-8\le -8-2\sqrt{23}. $$ Since it holds $\lim_{x\to-\infty}\left(x-2+\frac{23}{x+6}\right)=-\infty$, the range on $\{x+6<0\}$ is given by $$(-\infty,-8-2\sqrt{23}]$$ also by the intermediate value theorem.

Gathering them, the range of the given function is $$ (-\infty,-8-2\sqrt{23}]\cup[-8+2\sqrt{23},\infty). $$