Question $f(x) = \frac{-3x + 9}{x^2 - 8x + 15}$
Find the asymptotes, intercepts, and holes of the function $f(x)$ .

Studdy Solution

STEP 1

What is this asking? We need to find all the asymptotes, intercepts, and holes of a given function. Watch out! Don't forget to check for those sneaky holes – they can be easy to miss!
Also, remember the difference between vertical and horizontal asymptotes.

STEP 2

1. Simplify the function
2. Find vertical asymptotes
3. Find horizontal asymptotes
4. Find x-intercepts
5. Find y-intercepts
6. Find holes

STEP 3

We can factor out a $-3$ from the numerator: $-3x + 9 = -3(x - 3)$

STEP 4

We're looking for two numbers that multiply to $15$ and add to $-8$ .
Those numbers are $-3$ and $-5$ .
So, we can factor the denominator as: $x^2 - 8x + 15 = (x - 3)(x - 5)$

STEP 5

Now, we can rewrite the function $f(x)$ as: $f(x) = \frac{-3(x - 3)}{(x - 3)(x - 5)}$ We can divide the numerator and denominator by $(x-3)$ to get $1$ .
This gives us a simplified function: $f(x) = \frac{-3}{x - 5}$ Keep in mind that this simplification is only valid when $x \ne 3$ .

STEP 6

Vertical asymptotes occur when the denominator is equal to zero, and the numerator is not equal to zero.
In our simplified function, the denominator is $x - 5$ .

STEP 7

Setting the denominator equal to zero gives us: $x - 5 = 0$ Adding $5$ to both sides, we get $x = \mathbf{5}$ .
Since the numerator is $-3$ at $x=5$ , we have a vertical asymptote at $x = \mathbf{5}$ .

STEP 8

The degree of the numerator ( $0$ ) is less than the degree of the denominator ( $1$ ) in the simplified function.

STEP 9

When the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $y = \mathbf{0}$ .

STEP 10

To find x-intercepts, we set $f(x) = 0$ : $\frac{-3}{x - 5} = 0$

STEP 11

Multiplying both sides by $(x-5)$ , we get $-3 = 0$ , which is never true.
So, there are no x-intercepts.

STEP 12

To find y-intercepts, we set $x = 0$ in the simplified function: $f(0) = \frac{-3}{0 - 5}$

STEP 13

$f(0) = \frac{-3}{-5} = \frac{3}{5}$ So, the y-intercept is at $\left(0, \mathbf{\frac{3}{5}} \right)$ .

STEP 14

Remember when we simplified the function?
We cancelled out the $(x - 3)$ factor.

STEP 15

This cancellation indicates a hole at $x = 3$ .
To find the y-coordinate of the hole, plug $x = 3$ into the simplified function: $f(3) = \frac{-3}{3 - 5} = \frac{-3}{-2} = \frac{3}{2}$ So, there's a hole at $\left( \mathbf{3}, \mathbf{\frac{3}{2}} \right)$ .

STEP 16

Vertical Asymptote: $x = 5$ Horizontal Asymptote: $y = 0$ X-intercepts: None Y-intercept: $\left(0, \frac{3}{5} \right)$ Hole: $\left(3, \frac{3}{2} \right)$

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