Math  /  Algebra

QuestionMinga f(x)=2x+2x21f(x)=\frac{2 x+2}{x^{2}-1}
Answer Attempt 1 out of 2
Horizontal Asymptote: y=y= \square No horizontal asymptote
Vertical Asymptote: x=x= \square No vertical asymptote xx-Intercept: ( \square , 0) No xx-intercept \qquad yy-Intercept: ( 0, \square ) \qquad No yy-intercept
Hole: \square , \square No hole

Studdy Solution

STEP 1

What is this asking? We're on a mission to find the horizontal and vertical asymptotes, *x*-intercept, *y*-intercept, and any holes of the function f(x)=2x+2x21f(x) = \frac{2x + 2}{x^2 - 1}! Watch out! Don't forget to **simplify** the function first!
There might be hidden secrets lurking within.

STEP 2

1. Simplify the function
2. Find the vertical asymptotes
3. Find the horizontal asymptote
4. Find the *x*-intercept
5. Find the *y*-intercept
6. Find the hole

STEP 3

We can **factor out** a 22 from the numerator: 2x+2=2(x+1)2x + 2 = 2(x + 1).

STEP 4

The denominator is a **difference of squares**: x21=(x1)(x+1)x^2 - 1 = (x - 1)(x + 1).

STEP 5

Now, we have f(x)=2(x+1)(x1)(x+1)f(x) = \frac{2(x + 1)}{(x - 1)(x + 1)}.
Since (x+1)(x+1) appears in both the numerator and denominator, as long as x1x \ne -1, we can **divide** them out to get 11: x+1x+1=1\frac{x+1}{x+1} = 1.
So, our simplified function becomes f(x)=2x1f(x) = \frac{2}{x - 1} for x1x \ne -1.

STEP 6

Vertical asymptotes happen when the denominator is zero and the numerator isn't!
Setting the denominator of our simplified function to zero gives us x1=0x - 1 = 0.

STEP 7

Adding 11 to both sides gives us x=1x = 1.
This is our **vertical asymptote**!

STEP 8

The degree of the numerator (22) is less than the degree of the denominator (x21x^2 - 1 is degree 22).

STEP 9

When the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is always y=0y = 0.

STEP 10

To find the *x*-intercept, we set y=f(x)=0y = f(x) = 0, so we have 0=2x10 = \frac{2}{x - 1}.

STEP 11

There's no value of xx that will make this fraction equal to zero.
So, there's **no *x*-intercept**!

STEP 12

To find the *y*-intercept, we set x=0x = 0 in our simplified function: f(0)=201f(0) = \frac{2}{0 - 1}.

STEP 13

f(0)=21=2f(0) = \frac{2}{-1} = -2.
So, our *y*-intercept is (0,2)(0, -2).

STEP 14

Remember when we simplified the function and canceled out (x+1)(x + 1)?
That's where the hole is hiding!

STEP 15

Setting x+1=0x + 1 = 0 gives us x=1x = -1.

STEP 16

Substitute x=1x = -1 into the simplified function: f(1)=211=22=1f(-1) = \frac{2}{-1 - 1} = \frac{2}{-2} = -1.

STEP 17

So, the hole is at (1,1)(-1, -1).

STEP 18

Horizontal Asymptote: y=0y = 0 Vertical Asymptote: x=1x = 1 *x*-Intercept: None *y*-Intercept: (0,2)(0, -2) Hole: (1,1)(-1, -1)

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