QuestionFind the asymptotes, intercepts, and holes of the function $f(x) = \frac{5x^2 + 25x + 30}{3x + 6}$ .

Studdy Solution

STEP 1

What is this asking? We're on a mission to find out all the secrets of this function: where it explodes to infinity (asymptotes), where it crosses the axes (intercepts), and where it has sneaky little gaps (holes)! Watch out! Don't forget that holes happen when factors cancel out, and asymptotes happen when the denominator goes to zero but the numerator doesn't!

STEP 2

1. Simplify the function
2. Find the vertical asymptote
3. Find the x-intercept(s)
4. Find the y-intercept
5. Find the hole

STEP 3

Let's factor the numerator and denominator to see if we can simplify things!
We've got $5x^2 + 25x + 30$ .
We can factor out a $5$ to get $5(x^2 + 5x + 6)$ .
Now, we're looking for two numbers that add up to $5$ and multiply to $6$ .
Those numbers are $2$ and $3$ !
So, the numerator becomes $5(x+2)(x+3)$ .

STEP 4

The denominator is $3x + 6$ , and we can factor out a $3$ to get $3(x+2)$ .

STEP 5

Now, our function looks like this: $f(x) = \frac{5(x+2)(x+3)}{3(x+2)}$ We can divide to one the $(x+2)$ terms in the numerator and denominator, as long as $x \ne -2$ .
This gives us a simplified function: $f(x) = \frac{5(x+3)}{3}$ Keep in mind that this simplification is valid only when $x \ne -2$ .

STEP 6

Since we canceled out the $(x+2)$ term, there's no value of $x$ that makes the denominator of our simplified function zero.
This means there are no vertical asymptotes!

STEP 7

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

STEP 8

Multiplying both sides by $3$ gives us $5(x+3) = 0$ .
Dividing both sides by $5$ gives us $x+3 = 0$ .
Subtracting $3$ from both sides gives us $x = -3$ .
So, our x-intercept is $(-3, 0)$ .

STEP 9

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

STEP 10

$f(0) = \frac{5 \cdot 3}{3} = \frac{15}{3} = 5$ So, our y-intercept is $(0, 5)$ .

STEP 11

Remember that we canceled out the $(x+2)$ term earlier?
That's where our hole is!

STEP 12

The hole is at $x = -2$ .

STEP 13

To find the y-value, plug $x = -2$ into the simplified function: $f(-2) = \frac{5(-2+3)}{3} = \frac{5(1)}{3} = \frac{5}{3}$ So, our hole is at $(-2, \frac{5}{3})$ .

STEP 14

The function has an x-intercept at $(-3, 0)$ , a y-intercept at $(0, 5)$ , a hole at $(-2, \frac{5}{3})$ , and no vertical asymptotes!

Was this helpful?

Get Studdy

Scan QR code with your device to get Studdy on iOS or Android

QR code

Studdy solves anything!

Download the app

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Parents Influencer program Contact Policy Terms