Question

webwork 1 - topics 2 - 3: problem 15 (1 point) determine all horizontal asymptotes of the following function. recall that horizontal asymptotes are about long - term behaviour. in particular, it tries to answer the questions about how the function behaves as (x\toinfty) and (x\to-infty). (f(x)=\frac{sqrt{49x^{2}+2x + 6}-3x + 7}{6x + 1}) horizontal asymptotes at (y=) if there is more than one answer, enter them separated by commas. preview my answers submit answers you have attempted this problem 0 times. you have 6 attempts remaining. page generated september 20, 2025 at 2:19:28 pm mdt webwork © 1996 - 2025 | theme: math4 | ww_version: 2.20 | pg_version 2.20 the webwork project

Explanation:

Step1: Analyze as $x\to\infty$

Divide numerator and denominator by $x$. For the square - root part, $\sqrt{49x^{2}+2x + 6}=\vert x\vert\sqrt{49+\frac{2}{x}+\frac{6}{x^{2}}}$. As $x\to\infty$, $\vert x\vert=x$. So $f(x)=\frac{\sqrt{49x^{2}+2x + 6}-3x + 7}{6x + 1}=\frac{x\sqrt{49+\frac{2}{x}+\frac{6}{x^{2}}}-3x + 7}{6x + 1}$. Dividing by $x$ gives $\lim_{x\to\infty}\frac{\sqrt{49+\frac{2}{x}+\frac{6}{x^{2}}}-3+\frac{7}{x}}{6+\frac{1}{x}}$.

Step2: Evaluate the limit as $x\to\infty$

As $x\to\infty$, $\frac{2}{x}\to0$, $\frac{6}{x^{2}}\to0$, $\frac{7}{x}\to0$ and $\frac{1}{x}\to0$. Then $\lim_{x\to\infty}\frac{\sqrt{49+\frac{2}{x}+\frac{6}{x^{2}}}-3+\frac{7}{x}}{6+\frac{1}{x}}=\frac{\sqrt{49}-3}{6}=\frac{7 - 3}{6}=\frac{2}{3}$.

Step3: Analyze as $x\to-\infty$

As $x\to-\infty$, $\vert x\vert=-x$. So $\sqrt{49x^{2}+2x + 6}=-x\sqrt{49+\frac{2}{x}+\frac{6}{x^{2}}}$. Then $f(x)=\frac{-x\sqrt{49+\frac{2}{x}+\frac{6}{x^{2}}}-3x + 7}{6x + 1}$. Dividing by $x$ (and note that $\frac{-x}{x}=-1$), we get $\lim_{x\to-\infty}\frac{-\sqrt{49+\frac{2}{x}+\frac{6}{x^{2}}}-3+\frac{7}{x}}{6+\frac{1}{x}}$.

Step4: Evaluate the limit as $x\to-\infty$

As $x\to-\infty$, $\frac{2}{x}\to0$, $\frac{6}{x^{2}}\to0$, $\frac{7}{x}\to0$ and $\frac{1}{x}\to0$. Then $\lim_{x\to-\infty}\frac{-\sqrt{49+\frac{2}{x}+\frac{6}{x^{2}}}-3+\frac{7}{x}}{6+\frac{1}{x}}=\frac{-7 - 3}{6}=-\frac{5}{3}$.

Answer:

$y = \frac{2}{3},y=-\frac{5}{3}$