All categories

3 years ago

Consider the following equation f(x)=(x^2+4)/(4x^2-4x-8)

2 answers:

8 0

Answer: x=-1 and x=2 because (this is where the function is undefined) name of horizontal asymptote (y=1/4) because (m=n)

Step-by-step explanation: on edge

kifflom [539]3 years ago

3 0

Question:

Consider the following equation: f(x)=x^2+4\4x^2-4x-8 name the vertical asymptote(s)

Answer:

The vertical asymptotes are x = 2 or x = -1

Solution:

Given that,

$f(x)=\frac{x^2+4}{4x^2-4x-8}$

For vertical asymptotes, set the denominator to zero and then solve the quadratic equation

$4x^2-4x-8=0\\\\Divide\ by\ 2\\\\x^2 - 2x - 4 = 0$

$Split\ the\ middle\ term\\\\x^2+x-2x-2=0\\\\Group\ the\ equation\\\\(x^2 + x) - (2x + 2) = 0\\\\Factor\ the\ common\ term\\\\x(x + 1) - 2(x + 1) = 0\\\\(x + 1)(x - 2) = 0\\\\Therefore,\\\\x + 1 = 0\\\\x = -1\\\\And\\\\x - 2 = 0\\\\x = 2$

Thus the vertical asymptotes are x = - 1 and x = 2

You might be interested in

Which equation is true?

ch4aika [34]

Answer : $9x^2 - 25 = (3x - 5)(3x + 5)$

(1) $9x^2 - 25 = (3x - 5)(3x - 5)$

Use FOIL method to mulitply (3x – 5)(3x – 5)

$(3x - 5)(3x - 5) = 9x^2 - 30x + 25$

(2) $9x^2 - 25 = (3x - 5)(3x + 5)$

Use FOIL method to mulitply (3x - 5)(3x + 5)

$(3x - 5)(3x - 5) = 9x^2 - 15x + 15x - 25 = 9x^2 - 25$

(3) $9x^2 - 25 = -(3x + 5)(3x + 5)$

Use FOIL method to mulitply -(3x + 5)(3x + 5)

$-(3x + 5)(3x + 5) = -(9x^2 + 30x + 25) = -9x^2 -30x -25$

(4) $9x^2 - 25 = -(3x + 5)(3x - 5)$

Use FOIL method to mulitply -(3x + 5)(3x - 5)

$-(3x + 5)(3x - 5) = -(9x^2 - 25) = -9x^2 + 25$

So equation 2 is true

6 0

3 years ago

Read 2 more answers

Last year, there were 381 students at Woodland Middle School. This year, there are 419 students. What is the percent of change?

Nesterboy [21]

The is going to increase by 20

3 0

3 years ago

Describe any trends that you can see from the dot plot. Help

stealth61 [152]

Answer: Most planes flew below the optimal cruising speed. There is a cluster at the speeds 0.82 through 0.85. Also, the data is close to being symmetric

Step-by-step explanation:

4 0

3 years ago

g Suppose you are rasterizing a triangle with vertices (1, 1), (5, 1), and (2, 9), and the intensity values (e.g., redness) of t

Ivahew [28]

Answer:

6.333

Step-by-step explanation:

Our triangle vertices are

(1,1)(5,1)(2,9)

We are to perform interpolation at point(3,5)

x and y values are interpolated

On vertices (5,1)(2,9)

x = 3

x1 = 5

X2 = 2

Y1 = 1

Y2 = 9

Formula for interpolation

Y = [y1+(x-x1)(y2-y1)]/x2-x

Y = 1+(3-5)(9-1)/2-5

Y = 1+[-2*8/-3]

Y = 1+16/3

From here we use use LCM to get:

19/3 = 6.333

6 0

3 years ago

To evaluate log2(3), Autumn reasoned that since log2(2) = 1 and log2(4) = 2, log2(3) must be the average of 1 and 2 and therefor

geniusboy [140]

Answer:

log₂(3) = 1.585 ≠ 1.5

Her thinking is not valid because the technique of average is valid only if the graph of the function is a straight line, but the graph of the log function is not a straight line.

Therefore the values cannot be taken by average

Step-by-step explanation:

Given:

log₂(2) = 1

log₂(4) = 2

To evaluate : log₂(3)

Now,

we know that

logₓ(y) = $\frac{\log(y)}{\log(x)}$ (Here the log has same base in the numerator and the denominator i.e 10)

therefore,

log₂(3) = $\frac{\log(3)}{\log(2)}$

also,

log(2) = 0.3010

log(3) = 0.4771

thus,

log₂(3) = $\frac{0.4771}{0.3010}$

log₂(3) = 1.585 ≠ 1.5

Her thinking is not valid because the technique of average is valid only if the graph of the function is a straight line, but the graph of the log function is not a straight line.

Therefore the values cannot be taken by average

7 0

3 years ago