Find the intercepts, asymptotes (vertical, horizontal), and the location of any holes in the graph of F(x) = 5x^2-80/4x^2-40x+96. Then use that information to sketch a graph of F(x).

picture of question: https://drive.google.com/file/d/1qnZpkTbbws4FGhux870-I3acI7CleDeX/view?usp=sharing

Guest Jun 4, 2021

#1**+2 **

y-intercept is when x=0, so it is f(0), or -80/96 = -5/6 -> (0, -5/6)

x-intercept is when y=0, so it is when \(\frac{5x^2-80}{4x^2-40x+96}=0\). Knowing that x cannot equal the roots of \(4x^2-40x+96 \), namely, 4 and 6, we can multiply both sides in the original equation by \(4x^2-40x+96 \) having \(5x^2=80\), or \(x=4, -4\). From the previous part, however, x cannot equal 4, so the only x-intercept is (-4, 0).

For horizontal asymptotes, we cancel out factors \(\frac{5x^2-80}{4x^2-40x+96}=0 \) and use synthetic division to get \(5/4 +25/2(x-6)\). The larger x gets, the less \(25/2(x-6)\) is, meaning that the expression \(5/4 +25/2(x-6)\) gets closer and closer to 5/4. When x gets smaller, the less \(25/2(x-6)\) is, meaning that the expression approaches 5/4 but never reaches it. Thus, y = 5/4 is the horizontal asymptote.

For the vertical asymptote, we have x=4 and x=6 as possibilities (roots of \(4x^2-40x+96\), the denominator), though, since the (x-4) in the numerator and denominator cancel, x=4 is a hole in the graph while x=6 is a vertical asymptote. x=6 is the only vertical asymptote.

For the hole in the graph, we have x=4 (mentioned in the previous part), as it remains in the denominator and numerator, hence not allowing x=4, but still canceling during the graph overall. Putting in x=4, we have y = -5

y-intercept (0, -5/6)

x-intercept (-4, 0)

Horizontal asymptote y=5/4

Vertical asymptote x=6

Hole (4, -5)

The graph is available at desmos graphing calculator, though I do not have an account yet and thus cannot share it. It should be easy enough to create however. Anyways, this was a long response, but I believe that this answered the question. Have a great day!

EnchantedLava68 Jun 4, 2021

#2**+2 **

5x^2 - 80 5 ( x + 4)(x - 4) 5 (x +4)

______________ = _____________ = _______

4x^2 - 40x + 96 4 ( x - 6) (x - 4) 4 (x - 6)

The horizontal asymptote is y = 5/4

We have a vertical asymptote where

x - 6 = 0

x = 6

The y intercept = 20 / -24 = -5/6

The x intercept

x + 4 = 0

x = -4

We have a "hole" at

(x - 4) = 0

x = 4

Here's a graph : https://www.desmos.com/calculator/isyf32es2o

CPhill Jun 4, 2021