Math / Calculus

Question12. Determine the equation(s) of the vertical asymptote(s) of the function $y=\frac{0.9^{x}}{2 x^{2}-3}: \checkmark \checkmark \checkmark$

Studdy Solution

STEP 1

What is this asking? Where does this function explode to infinity because the denominator becomes zero? Watch out! Don't forget to check if the numerator is also zero at the same $x$ values.
If both numerator and denominator are zero, it might not be a vertical asymptote!

STEP 2

1. Find suspicious points
2. Check for holes

STEP 3

To find the vertical asymptotes, we need to find the $x$ values where the denominator is zero.
So, let's set the denominator equal to zero and solve for $x$ : $2x^2 - 3 = 0$

STEP 4

Let's isolate the $x^2$ term by adding $3$ to both sides: $2x^2 = 3$

STEP 5

Now, divide both sides by $2$ : $x^2 = \frac{3}{2}$ Finally, take the square root of both sides: $x = \pm\sqrt{\frac{3}{2}}$ So, our suspicious points are $x = \sqrt{\frac{3}{2}}$ and $x = -\sqrt{\frac{3}{2}}$ .

STEP 6

Now, let's make sure the numerator isn't zero at these suspicious points.
If the numerator is zero, we might have a hole instead of an asymptote.
Let's check!
The numerator is $0.9^x$ .
Since $0.9$ raised to any power will never be zero, we're in the clear!
No holes here!

STEP 7

The equations of the vertical asymptotes are $x = \sqrt{\frac{3}{2}}$ and $x = -\sqrt{\frac{3}{2}}$ .
Boom!

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Question12. Determine the equation(s) of the vertical asymptote(s) of the function $y=\frac{0.9^{x}}{2 x^{2}-3}: \checkmark \checkmark \checkmark$

Studdy Solution

STEP 1

What is this asking? Where does this function explode to infinity because the denominator becomes zero? Watch out! Don't forget to check if the numerator is also zero at the same $x$ values.
If both numerator and denominator are zero, it might not be a vertical asymptote!

STEP 2

1. Find suspicious points
2. Check for holes

STEP 3

To find the vertical asymptotes, we need to find the $x$ values where the denominator is zero.
So, let's set the denominator equal to zero and solve for $x$ : $2x^2 - 3 = 0$

STEP 4

Let's isolate the $x^2$ term by adding $3$ to both sides: $2x^2 = 3$

STEP 5

Now, divide both sides by $2$ : $x^2 = \frac{3}{2}$ Finally, take the square root of both sides: $x = \pm\sqrt{\frac{3}{2}}$ So, our suspicious points are $x = \sqrt{\frac{3}{2}}$ and $x = -\sqrt{\frac{3}{2}}$ .

STEP 6

Now, let's make sure the numerator isn't zero at these suspicious points.
If the numerator is zero, we might have a hole instead of an asymptote.
Let's check!
The numerator is $0.9^x$ .
Since $0.9$ raised to any power will never be zero, we're in the clear!
No holes here!

STEP 7

The equations of the vertical asymptotes are $x = \sqrt{\frac{3}{2}}$ and $x = -\sqrt{\frac{3}{2}}$ .
Boom!

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Question12. Determine the equation(s) of the vertical asymptote(s) of the function y=0.9x2x2−3:✓✓✓y=\frac{0.9^{x}}{2 x^{2}-3}: \checkmark \checkmark \checkmarky=2x2−30.9x​:✓✓✓

Studdy Solution

STEP 1

What is this asking? Where does this function explode to infinity because the *denominator* becomes zero? Watch out! Don't forget to check if the *numerator* is also zero at the same xxx values.If both numerator and denominator are zero, it might not be a vertical asymptote!

STEP 2

1. Find suspicious points2. Check for holes

STEP 3

To find the vertical asymptotes, we need to find the xxx values where the denominator is zero.So, let's **set the denominator equal to zero** and **solve for** xxx: 2x2−3=02x^2 - 3 = 02x2−3=0

STEP 4

Let's **isolate the** x2x^2x2 term by **adding** 333 to both sides: 2x2=32x^2 = 32x2=3

STEP 5

Now, **divide both sides** by 222: x2=32x^2 = \frac{3}{2}x2=23​ Finally, **take the square root of both sides**: x=±32x = \pm\sqrt{\frac{3}{2}}x=±23​​ So, our **suspicious points** are x=32x = \sqrt{\frac{3}{2}}x=23​​ and x=−32x = -\sqrt{\frac{3}{2}}x=−23​​.

STEP 6

Now, let's make sure the numerator isn't zero at these **suspicious points**.If the numerator *is* zero, we might have a hole instead of an asymptote.Let's check!The numerator is 0.9x0.9^x0.9x.Since 0.90.90.9 raised to *any* power will *never* be zero, we're in the clear!No holes here!

STEP 7

The equations of the vertical asymptotes are x=32x = \sqrt{\frac{3}{2}}x=23​​ and x=−32x = -\sqrt{\frac{3}{2}}x=−23​​.Boom!

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Question12. Determine the equation(s) of the vertical asymptote(s) of the function $y=\frac{0.9^{x}}{2 x^{2}-3}: \checkmark \checkmark \checkmark$

What is this asking? Where does this function explode to infinity because the denominator becomes zero? Watch out! Don't forget to check if the numerator is also zero at the same $x$ values.
If both numerator and denominator are zero, it might not be a vertical asymptote!

1. Find suspicious points
2. Check for holes

To find the vertical asymptotes, we need to find the $x$ values where the denominator is zero.
So, let's set the denominator equal to zero and solve for $x$ : $2x^2 - 3 = 0$

Let's isolate the $x^2$ term by adding $3$ to both sides: $2x^2 = 3$

Now, divide both sides by $2$ : $x^2 = \frac{3}{2}$ Finally, take the square root of both sides: $x = \pm\sqrt{\frac{3}{2}}$ So, our suspicious points are $x = \sqrt{\frac{3}{2}}$ and $x = -\sqrt{\frac{3}{2}}$ .

Now, let's make sure the numerator isn't zero at these suspicious points.
If the numerator is zero, we might have a hole instead of an asymptote.
Let's check!
The numerator is $0.9^x$ .
Since $0.9$ raised to any power will never be zero, we're in the clear!
No holes here!

The equations of the vertical asymptotes are $x = \sqrt{\frac{3}{2}}$ and $x = -\sqrt{\frac{3}{2}}$ .
Boom!