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find \(q(x)\) if the graph of \(\frac{4x-x^3}{q(x)} \) has a hole at \(x=-2\), a vertical asymptote at \(x=1\), no horizontal asymptote, and \(q(3)=-30\)

 

 

much help appreciated! :((( 

 Mar 23, 2020
 #1
avatar+485 
+1

First, we observe that if the function \(q(x)\) has a hole at x = -2, that would indicate that both the values of the numerator AND denominator are 0. (0/0 produces the hole in the graph). The important takeway here is that the function \(q(x)\) then has a root of -2

 

If the equation \(q(x)\) has a vertical asymptote at x = 1, that would indicate that the solution present at 1 is undefined. We know from this that 1 is a root of \(q(x)\)

 

Since we have 2 roots of \(q(x)\) let's assume it's a quadratic with two roots(anyone have a more rigorous proof than this?). We have the general form as:

 

ax2+bx+c = 0

We know that -2 and 1 are roots of this quadratic, so by vietas formulas:

 

-2 + 1 = -b/a

-1 = -b/a

-a = -b

a = b

 

-2 * 1 = c /a 

-2 = c/a

c = -2a

(we will use these later :P)

Next we look at the final thing the problem gives us, which is that \(f(3) = 30\)

That means that substituting into the expression, we get:

 

9a + 3b + c = 30

 

Here's where our previous expressions come in handy:

 

We can express everything in terms of b here. We then get:

 

9(b) + 3b + c = 30

 

c = -2a

a = b

c = -2b

9b + 3b -2b = 03

10b = 30

b = 3

 

Since a = b

a = 3

 

since c = -2b:

c = -6

 

we then have the quadratic:

 

3x2+3x-6

on the bottom

 Mar 24, 2020
edited by jfan17  Mar 24, 2020
 #2
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That is really impressive, jfan! Great job!!!!

CalTheGreat  Mar 24, 2020
 #3
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i entered my answer (for q(x)) as 3x^2+3x-6 and it was wrong... :((( i still have one more attempt...

 Mar 24, 2020
 #4
avatar+485 
+1

Oh sorry, I saw your question as q(3) = 30, not -30. I already showed you how I would've gotten it with f(3) = 30, so can you build off of that?

jfan17  Mar 24, 2020
edited by jfan17  Mar 24, 2020
 #5
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+1

thanks!! i just changed it to -30 and got the correct answer. you're amazing :D

 Mar 24, 2020
 #6
avatar+485 
+1

No problem!

jfan17  Mar 24, 2020

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