Question

Consider the following rational function:
y=2x+5x+1
Write down the coordinates of the x-intercept & y-intercept.
State the equations of the horizontal & vertical asymptotes.
Sketch the graph of the function clearly labelling the axis intercepts & the asymptotes.

Answer 1

Answer:

By looking at the graph of a rational function, we can investigate its local behavior and easily see whether there are asymptotes. We may even be able to approximate their location. Even without the graph, however, we can still determine whether a given rational function has any asymptotes, and calculate their location.

Vertical Asymptotes

The vertical asymptotes of a rational function may be found by examining the factors of the denominator that are not common to the factors in the numerator. Vertical asymptotes occur at the zeros of such factors.

HOW TO: GIVEN A RATIONAL FUNCTION, IDENTIFY ANY VERTICAL ASYMPTOTES OF ITS GRAPH.

Factor the numerator and denominator.

Note any restrictions in the domain of the function.

Reduce the expression by canceling common factors in the numerator and the denominator.

Note any values that cause the denominator to be zero in this simplified version. These are where the vertical asymptotes occur.

Note any restrictions in the domain where asymptotes do not occur. These are removable discontinuities.

EXAMPLE 5: IDENTIFYING VERTICAL ASYMPTOTES

Find the vertical asymptotes of the graph of \displaystyle k\left(x\right)=\frac{5+2{x}^{2}}{2-x-{x}^{2}}k(x)=

2−x−x

2

5+2x

2

.

SOLUTION

First, factor the numerator and denominator.

⎧⎪⎨⎪⎩

(x)=5+2x22−x−x2=5+2x2(2+x)(1−x)

To find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero:

{(2+x)(1−x)=0x=−2,1

Neither \displaystyle x=-2x=−2 nor \displaystyle x=1x=1 are zeros of the numerator, so the two values indicate two vertical asymptotes. Figure 9 confirms the location of the two vertical asymptotes.

Step-by-step explanation:

hope it's helpful to you

mark me as brainliests

Answer 2

Answer: 2x+5=7x

               7x+1=8x

Step-by-step explanation: SOLUTION

First, factor the numerator and denominator.

⎧⎪⎨⎪⎩

(x)=5+2x22−x−x2=5+2x2(2+x)(1−x)

To find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero:

{(2+x)(1−x)=0x=−2,1

Neither \displaystyle x=-2x=−2 nor \displaystyle x=1x=1 are zeros of the numerator, so the two values indicate two vertical asymptotes. Figure 9 confirms the location of the two vertical asymptotes. mark me as brainliest PLEASEEE!!!! MARK me as BRAINLIEST ;(