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2 years ago

What is the horizontal asympote of this graph?

Mathematics

1 answer:

Anna35 [415]2 years ago

4 0

Using it's concept, it is found that the graph has no horizontal asymptote.

<h3>What are the horizontal asymptotes of a function f(x)?</h3>

The horizontal asymptote is the value of f(x) as x goes to infinity, as long as this value is different of infinity.

In this problem, we have that:

The function is undefined for x < 0, hence $\lim_{x \rightarrow -\infty} f(x)$ is undefined.

For x > 0, the funciton goes to infinity, hence $\lim_{x \rightarrow \infty} f(x) = \infty$ .

Thus, the graph has no horizontal asymptote.

More can be learned about horizontal asymptotes at brainly.com/question/16948935

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You might be interested in

The length of segment XY is 9 cm. Which statements regarding triangle XYZ are correct? Check all that apply.

Dominik [7]

<u>Answer-</u>

YZ = 9 cm
$XZ=9\sqrt2$
XZ is the longest segment in triangle XYZ.

<u>Solution-</u>

Triangle XYZ is a right angle triangle with ∠Y=90°

Triangle XYZ is also an isosceles triangle, as

∠X =45°= ∠Z

Therefore, triangle XYZ is a right angle isosceles triangle.

So,

XY = YZ = 9 cm

Applying Pythagoras theorem,

$\Rightarrow XZ^2=XY^2+YZ^2\\\\\Rightarrow XZ^2=9^2+9^2\\\\\Rightarrow XZ^2=2\times 9^2\\\\\Rightarrow XZ=\sqrt{2\times 9^2}=9\sqrt2$

In a right angle triangle, always the hypotenuse is the largest. So XZ is the longest segment in triangle XYZ.

6 0

3 years ago

Read 2 more answers

Find the range and interquartile range of the data.

shutvik [7]

I believe the range is the middle number so start from 63 then 46 then 53 then go back and forth until you get the middle number now after you get the middle number for the IQR add up both sides of the middle and / by the all the number on the left and right of the middle number

4 0

2 years ago

Find the solution to the equation below.<br> 2x^2+3x-20=0

mariarad [96]

Answer:

x = -4, 5/2

Step-by-step explanation:

A quadratic can be solved may ways, including graphing, factoring, and the quadratic formula. You can also check possible answers by making use of the relationships between solutions and the coefficients.

A graph is attached. It shows the solutions to be -4 and 5/2.

When factored, the equation becomes ...

(2x -5)(x +4) = 0 . . . . . has solutions x=-4, x=5/2 (these make the factors zero)

Using the quadratic formula, the solutions of ax^2 +bx +c = 0 are found from ...

x = (-b±√(b²-4ac))/(2a)

x = (-3±√(3²-4(2)(-20))/(2(2)) = (-3±√169)/4 = {-16, +10}/4

x = {-4, 5/2}

For ax^2 +bx +c = 0, the solutions must satisfy ...

product of solutions is c/a = -20/2 = -10

Only the first and last choices have this product.

sum of solutions is -b/a = -3/2

Only the first choice (-4, 5/2) has this sum.