What is the horizontal asymptote of this graph? ( need answer asap)

What is the horizontal asymptote of this graph?

y = 0

Q1:    

Q2:    

Q3:    

Q4:    

Step-by-step explanation:

If there is a vertical asymptote at x=k, then the denominator must contain a term (x-k) or (x-k) raised to a positive integer power.

If there is an x intercept at k, then the numerator must contain a term (x-k) or (x-k) raised to a positive integer power.

If there is a horizontal asymptote of 0, then the highest power of x in the denominator must exceed the highest power of x in the numerator.

If thee is a horizontal asymptote of k, then the highest power of x in the numerator and denominator must be the same and the ratio of their coefficients (numerator to denominator) must be k.

Q1:    

Q2:    

Q3:    

Q4:    

Step-by-step explanation:

If there is a vertical asymptote at x=k, then the denominator must contain a term (x-k) or (x-k) raised to a positive integer power.

If there is an x intercept at k, then the numerator must contain a term (x-k) or (x-k) raised to a positive integer power.

If there is a horizontal asymptote of 0, then the highest power of x in the denominator must exceed the highest power of x in the numerator.

If thee is a horizontal asymptote of k, then the highest power of x in the numerator and denominator must be the same and the ratio of their coefficients (numerator to denominator) must be k.

As shown in picture,

Option 2 is correct.

Horizontal asymptote at y = 0 (graph approaches to 0 when x at -infinity and + infinity)

Option 4 is correct.

Vertical asymptote at x = 6 ( graph goes to infinity when x = 6)

Option 5 is correct.

Vertical asymptote at x = -4 ( graph goes to infinity when x = -4)

Hope this helps!

:)

As shown in picture,

Option 2 is correct.

Horizontal asymptote at y = 0 (graph approaches to 0 when x at -infinity and + infinity)

Option 4 is correct.

Vertical asymptote at x = 6 ( graph goes to infinity when x = 6)

Option 5 is correct.

Vertical asymptote at x = -4 ( graph goes to infinity when x = -4)

Hope this helps!

:)

y = -2(x+2)(x-1)/((x+3)(x-6)) = (-2x^2 -2x +4)/(x^2 -3x -18)

Step-by-step explanation:

A polynomial function will have a zero at x=a if it has a factor of (x-a). For the rational function to have zeros at x=-2 and x=1, the numerator factors must include (x+2) and (x-1).

For the function to have vertical asymptotes at x=-3 and x=6, the denominator of the rational function must have zeros there. That is, the denominator must have factors (x+3) and (x-6). Then the function with the required zeros and vertical asymtotes must look like ...

f(x) = (x+2)(x-1)/((x+3)(x-6))

This function will have a horizontal asymptote at x=1 because the numerator and denominator degrees are the same. In order for the horizontal asymptote to be -2, we must multiply this function by -2.

The rational function may be ...

y = -2(x +2)(x -1)/((x +3)(x -6))

If you want the factors multiplied out, this becomes

y = (-2x^2 -2x +4)/(x^2 -3x -18)

Given vertical asymptotes x = 2 and x = -5 .

Therefore Factors of denominator (x-2)(x+5).

x-intercepts (-1,0) and (1,0) .

Therefore factors of (x+1)(x-1).

So far we got rational function = .

Now, we are given horizontal asymptote at y = 9.

Multiplying above express by 9 on the top, we get

.

The rational function is

The horizontal asymptote occur at y = 1

Step-by-step explanation:

Given that the function has an x-intercept at (-9, 0), we have;

(x + 9) is a factor of the function

Also, given that the function has a vertical asymptote at x = 3, we have that a factor of the denominator of the function is (x - 3)

With a hole at (1, -5), we have, (x - 1) in both the numerator and the denominator and when x = 1, f(x) = -5

The general form of the function can be presented as follows;

f(x) = ((x + 9)·(x - 1))/((x - 3)(x - 1)) = (x² + 8·x - 9)/(x² - 4·x + 3)

Given that the power of the numerator and the denominator are equal, the horizontal asymptote occur at ratio of the leading coefficient of the numerator and the denominator which from the above equation is 1/1 = 1

The horizontal asymptote occur at y = 1

Please find attached the graph of the function created with Microsoft Excel

The rational function is

The horizontal asymptote occur at y = 1

Step-by-step explanation:

Given that the function has an x-intercept at (-9, 0), we have;

(x + 9) is a factor of the function

Also, given that the function has a vertical asymptote at x = 3, we have that a factor of the denominator of the function is (x - 3)

With a hole at (1, -5), we have, (x - 1) in both the numerator and the denominator and when x = 1, f(x) = -5

The general form of the function can be presented as follows;

f(x) = ((x + 9)·(x - 1))/((x - 3)(x - 1)) = (x² + 8·x - 9)/(x² - 4·x + 3)

Given that the power of the numerator and the denominator are equal, the horizontal asymptote occur at ratio of the leading coefficient of the numerator and the denominator which from the above equation is 1/1 = 1

The horizontal asymptote occur at y = 1

Please find attached the graph of the function created with Microsoft Excel

The correct option is;

The graph has a vertical asymptote at x = -3

Step-by-step explanation:

Whereby the given function is ,  we have;

The function is undefined at x = -3, therefore, the function has a verical asymptote at x = -3

The graph crosses the x-axis (the x-intercept) when the y coordinate, value h(x) = 0, therefore, we have;

Which gives;

2·x² + 6·x - 6 = 0

Dividing by 2 gives;

x² + 3·x - 3 = 0

x = (-3 ± √(3² - 4×1×(-3)))/(2 × 1) = (-3 ± √21)/2

Similarly, the y-intercept occurs when h(0) = -2 as follows;

Therefore, the graph crosses the y-axis, the y-intercept at (0, -2)

The correct option is, that the graph has a vertical asymptote at x = -3