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fomenos
3 years ago
5

Which function has a graph with a vertical

Mathematics
2 answers:
raketka [301]3 years ago
6 0

Answer:

f(x)=\frac{1}{x-7}-3, see the graph attached for a visual reference.

Step-by-step explanation:

Vertical asymptotes are only present in rational functions where the parent function f(x)=\frac{1}{x} has a vertical asymptote at the line x=0 and a horizontal asymptote at the line y=0. Because the vertical asymptote has to be x=7, the denominator must be x-7 in order for the denominator to equal 0. For the horizontal asymptote to be y=-3, then 3 must be subtracted from the rational function. Therefore, the function that has these asymptotes is f(x)=\frac{1}{x-7}-3.

Snezhnost [94]3 years ago
5 0
F(x) = 3 / (-3x - 3)

To find the vertical asymptotes, set the denominator of f(x) equal to zero.

-3x - 3 = 0

-3x = 3
x = -1

The vertical asymptote is
x = -1


To find the horizontal asymptotes, we examine the degree of the numerator and denominator of f(x). We use the leading terms to determine the degree. The leading term in the numerator is 7x0. The leading term of the denominator is -3x1. Since the exponent of x in the numerator is 0, the degree is 0 . Since the exponent of x in the denominator is 1, the degree is 1. The degree in the numerator is less than the degree in the denominator. Therefore, the horizontal asymptote is

y = 0

This also indicates that there are no x-intercepts.

To find the y-intercepts, set x=0.

The y-intercept is (0, -7/3).
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Answer:

The value of k is 4 and q is 16.

Step-by-step explanation:

First, you have to expand the brackets by applying :

{(a + b)}^{2}  =  {a}^{2}  + 2ab +  {b}^{2}

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{x}^{2}  + 8x + q =  {x}^{2}  + 2kx +  {k}^{2}

Next, you have to do it by comparison :

{x}^{2}  + 8x + q =  {x}^{2}  + 2kx +  {k}^{2}

by \: comparison

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k = 4

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Step-by-step explanation:

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