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# rational functions

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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

Jun 4, 2021

#1
+122
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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!

Jun 4, 2021
#2
+119771
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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

Jun 4, 2021