Guestion 14 of 49 (I point) I Question Attempt: 1 of Unilmited
For ,
(a) Identify the horizontal asymptotes (if any).
(b) If the graph of the function has a horizontal asymptote, determine the point (if any) where the graph crosses the horizontal asymptote(s).
Separate multiple equations of asymptotes with commas as necessary. Select “None” if applicable.

Ask by Marsh Davison. in the United States

Nov 09,2024

Question

Guestion 14 of 49 (I point) I Question Attempt: 1 of Unilmited
For ,
(a) Identify the horizontal asymptotes (if any).
(b) If the graph of the function has a horizontal asymptote, determine the point (if any) where the graph crosses the horizontal asymptote(s).
Separate multiple equations of asymptotes with commas as necessary. Select “None” if applicable.

Ask by Marsh Davison. in the United States

Nov 09,2024

UpStudy AI · Accepted Answer

Find the horizontal asymptotes of $$ (5*x-2)/(2*x^2+4*x-4) $$.
Function by following steps:
- step0: Find the horizontal asymptotes:
  $$y=\frac{5x-2}{2x^{2}+4x-4}$$
- step1: Evaluate the limits $$\lim _{xightarrow +\infty}\left(yight)$$ and $$\lim _{xightarrow -\infty}\left(yight):$$
  $$\begin{align}&\lim _{xightarrow +\infty}\left(\frac{5x-2}{2x^{2}+4x-4}ight)\&\lim _{xightarrow -\infty}\left(\frac{5x-2}{2x^{2}+4x-4}ight)\end{align}$$
- step2: Calculate:
  $$\begin{align}&0\&0\end{align}$$
- step3: The finite values are horizontal asymptotes:
  $$\begin{align}&y=0\end{align}$$
The horizontal asymptote of the function $$ I(x)=\frac{5x-2}{2x^{2}+4x-4} $$ is $$ y=0 $$.

Since the horizontal asymptote is $$ y=0 $$, the graph of the function crosses the horizontal asymptote at the point where the function intersects the x-axis. The point of intersection is where the function equals zero.

To find the point where the graph crosses the horizontal asymptote, we need to solve the equation $$ I(x) = 0 $$. Let's solve this equation.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\frac{\left(5x-2ight)}{\left(2x^{2}+4x-4ight)}=0$$
- step1: Find the domain:
  $$\frac{\left(5x-2ight)}{\left(2x^{2}+4x-4ight)}=0,x \in \left(-\infty,-\sqrt{3}-1ight)\cup \left(-\sqrt{3}-1,\sqrt{3}-1ight)\cup \left(\sqrt{3}-1,+\inftyight)$$
- step2: Remove the parentheses:
  $$\frac{5x-2}{2x^{2}+4x-4}=0$$
- step3: Cross multiply:
  $$5x-2=\left(2x^{2}+4x-4ight)	imes 0$$
- step4: Simplify the equation:
  $$5x-2=0$$
- step5: Move the constant to the right side:
  $$5x=0+2$$
- step6: Remove 0:
  $$5x=2$$
- step7: Divide both sides:
  $$\frac{5x}{5}=\frac{2}{5}$$
- step8: Divide the numbers:
  $$x=\frac{2}{5}$$
- step9: Check if the solution is in the defined range:
  $$x=\frac{2}{5},x \in \left(-\infty,-\sqrt{3}-1ight)\cup \left(-\sqrt{3}-1,\sqrt{3}-1ight)\cup \left(\sqrt{3}-1,+\inftyight)$$
- step10: Find the intersection:
  $$x=\frac{2}{5}$$
The point where the graph of the function crosses the horizontal asymptote $$ y=0 $$ is at $$ x=\frac{2}{5} $$ or $$ x=0.4 $$.
Horizontal asymptote: $$ y=0 $$, crosses at $$ x=\frac{2}{5} $$ or $$ x=0.4 $$.

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