Question

Let $f(x) = \frac{2x^2 + 5x - 3}{3x^2 - 20x + 25}$ \newline This function has: \newline 1) A y intercept at the point \newline 2) x intercepts at the point(s) \newline 3) Vertical asymptotes at x =

          Let $f(x) = \frac{2x^2 + 5x - 3}{3x^2 - 20x + 25}$ \newline This function has: \newline 1) A y intercept at the point \newline 2) x intercepts at the point(s) \newline 3) Vertical asymptotes at x =
        
Let f(x) = (2x^2 + 5x - 3)/(3x^2 - 20x + 25) This function has: 1) A y intercept at the point 2) x intercepts at the point(s) 3) Vertical asymptotes at x =

Added by Clayton C.

Close

Precalculus with Limits
Precalculus with Limits
Ron Larson 2nd Edition
AceChat toggle button
Close icon
Ace pointing down

Please give Ace some feedback

Your feedback will help us improve your experience

Thumb up icon Thumb down icon
Thanks for your feedback!
Profile picture
Let f(x)=(2x^(2)+5x-3)/(3x^(2)-20x+25) This function has: Ay intercept at the point x intercepts at the point(s Vertical asymptotes at x= 22+5x3 Let f() 322025 This function has: 1)A y intercept at the point 2) x intercepts at the point(s 3) Vertical asymptotes at x
Close icon
Play audio
Feedback
Powered by NumerAI
Danielle Fairburn David Collins
Kathleen Carty verified

Tim Thornhill and 80 other subject Precalculus educators are ready to help you.

Ask a new question

*

Labs

-

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

*

Key Concepts

-
Key Concept
Premium Feature
Explore the core concept behind this problem.
Play button
Key Concept
Premium Feature
Explore the core concept behind this problem.
Your browser does not support the video tag.

*

Recommended Videos

-
let-f-x-2-x-2-11-x-15-2-x-2-7-x-4-this-function-has-1-a-y-intercept-at-the-point-2-x-intercepts-at-the-points-3-vertical-asymptotes-at-x

Let f ( x ) = 2 x 2 − 11 x + 15 2 x 2 − 7 x − 4 . This function has: 1) A y-intercept at the point 2) x-intercepts at the point(s) 3) Vertical asymptotes at x =

Tim T.

let-f-x-3-x-2-7-x-6-2-x-2-3-x-1-this-function-has-1-a-y-intercept-at-the-point-2-x-intercepts-at-the-points-3-vertical-asymptotes-at-x

Let f ( x ) = 3 x 2 + 7 x − 6 2 x 2 − 3 x + 1 . This function has: 1) A y-intercept at the point 2) x-intercepts at the point(s) 3) Vertical asymptotes at x =

Donna D.

let-f-x-3-x-2-7-x-6-2-x-2-3-x-1-this-function-has-1-a-y-intercept-at-the-point-2-x-intercepts-at-the-points-3-vertical-asymptotes-at-x-4-horizontal-asymptote-at-y

Let f ( x ) = 3 x 2 + 7 x − 6 2 x 2 − 3 x + 1 . This function has: 1) A y-intercept at the point 2) x-intercepts at the point(s) 3) Vertical asymptotes at x = 4) Horizontal asymptote at y =

Tim T.


*

Recommended Textbooks

-
Precalculus with Limits

Precalculus with Limits

Ron Larson 2nd Edition
achievement 1,164 solutions
Precalculus

Precalculus

Robert Blitzer 5th Edition
achievement 1,688 solutions
Precalculus

Precalculus

Jay Abramson 1st Edition
achievement 1,519 solutions

*

Transcript

-
00:01 This problem gives us the function f of x equals 2x squared minus 11x plus 15 over 2x squared minus 7x minus 4.
00:07 And we want to find the y -intercept for this function, the x -intercepts, and the vertical asymptotes for this rational function.
00:13 And first for the y -intercept, that is a point where we always have zero for the x value, and then whatever y value we're crossing the y -axis at.
00:20 And we could evaluate our function at zero, but there's a trick about rational functions where you can just look at the constants at the end of your function and just take the ratio of those, because when we plug in zero for all of our x values, zero squared will be zero, and zero times anything will be zero, so we're just going to be left with 15 over negative 4.
00:42 So that would be zero, negative 15, fourths as our y -intercept.
00:47 Now to find our x -intercepts as well as our vertical asymptotes, we're going to do something very similar, but we're going to do it with the numerator and denominator depending on which one we're trying to find.
00:56 If we're trying to find x -intercepts, that occurs when our numerator turns to zero, because zero divided by a value is still zero.
01:03 So we can solve for these x -intercepts by factoring, and we can't factor out our 2, so we have to multiply this a value times c, which gives us positive 30.
01:11 And then we ask ourselves what two numbers multiply to be positive 30, but add to be negative 11.
01:15 And those two numbers are negative 6 and negative 5.
01:19 But since we multiplied by a to start our factoring, we have to divide by a now, and that gives us the factored form of x minus 3, and then 2x minus 5, after 5 halves won't simplify and we have to take the 2 in front.
01:33 So when this factored form is set equal to zero, we get two solutions...
Need help? Use Ace
Ace is your personal tutor. It breaks down any question with clear steps so you can learn.
Start Using Ace
Ace is your personal tutor for learning
Step-by-step explanations
Instant summaries
Summarize YouTube videos
Understand textbook images or PDFs
Study tools like quizzes and flashcards
Listen to your notes as a podcast
Continue solving this problem
Create a free account to:
  • View full step-by-step solution
  • Ask follow-up questions with Ace AI
  • Save progress and study later
Continue Free
Join the community

18,000,000+

Students on Numerade


Trusted by students at 8,000+ universities

Numerade

Get step-by-step video solution
from top educators

Continue with Clever
or



By creating an account, you agree to the Terms of Service and Privacy Policy
Already have an account? Log In

A free answer
just for you

Watch the video solution with this free unlock.

Numerade

Log in to watch this video
...and 100,000,000 more!


EMAIL

PASSWORD

OR
Continue with Clever