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I need to find the range of $\frac{x^2+6}{2x+1}$

I know that $x$ cannot be $-1/2$. I graphed the function on Desmos and I can see that there is a vertical asymptote at $x=-1/2$ However, I'm having trouble finding the range of this function?

I can understand what the range is by looking at the graph however I don't know how to find this algebraically. Is there a method that I can use every time to get the range?

Also I know there is an oblique asymptote at $1/2x-1/4$. How can I use this to help me?

The range according to the graph is $(-\infty,-3)\cup(2,\infty)$

user8290579
  • 886

3 Answers3

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Here's a method without calculus.

Let $f(x)=y=\frac{x^2+6}{2x+1}\Rightarrow 2xy+y=x^2+6\Rightarrow x^2-2xy+6-y=0\Rightarrow x=\frac{2y±\sqrt{4y^2-24+4y}}{2}$ Denominator can never be $0$ in this case. So the domain depends only upon $4y^2-24+4y\Rightarrow 4y^2+4y-24≥0\Rightarrow 4(y-2)(y+3)≥0\Rightarrow (y-2)(y+3)≥0\Rightarrow y\in(- \infty, -3]\cup[2,\infty).$

Hence the range of the $f(x)$ is $(- \infty, -3]\cup[2,\infty)$

I guess you know solving inequalities...

An Alien
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To find the range of a function, my first instinct is to check whether the graph has an inverse. We can do this informally using the Horizontal Line Test. If no Horizontal Line intersects the function more than once, then the function has an inverse. With that, we can find $f(y) = x$. The "domain" of $f(y)$ would be the range of the function $f(x)$.

Else, we can use the derivatives of the function to help us find the maximum and minimum which gives us the lower and upper bound of the range of $f(x)$.

user3497
  • 43
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Using the resource shared by @user10101, we have that:

$$y=\dfrac{x^2+6}{2x+1}\Rightarrow y(2x+1)=x^2+6\Rightarrow 0=x^2-2xy-y+6$$

Then we can use the quadratic formula, which yields: $$x=\dfrac{2y\pm\sqrt{4y^2+4y-24}}{2}=y\pm\sqrt{y^2+y-6}=y\pm\sqrt{(y+3)(y-2)}$$

Since the term inside the square root has to be nonnegative, we have that if $y\geq2$ then the term is always positive. On the other hand, if $y<2$ then $(y-2)$ is negative, so $(y+3)$ has to be negative to get a positive sign, which happens when $y\leq-3$. That gives you the range you obtained.

Schach21
  • 700