+0

0
53
6

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
+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

3x2+3x-6

on the bottom

Mar 24, 2020
edited by jfan17  Mar 24, 2020
#2
+1

That is really impressive, jfan! Great job!!!!

CalTheGreat  Mar 24, 2020
#3
0

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
+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
+1

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

Mar 24, 2020
#6
+1

No problem!

jfan17  Mar 24, 2020