Question

@UnkleRhaukus

Answer 1

The hole in the graph, if any, would be at:


Select one:


a. x = 3

b. x = –3

c. y = 3

d. No holes in the graph

Answer 2

First, you would have to find where either \(x\) or \(y\) is a complex number. You can do that by substituting either:\[x=3\]\[x=-3\]\[y=3\] And see which gives a complex number

Answer 3

I'll help for one: \(x=3\)
Equation: \[y=\frac{4x^2+12x}{x^2-x+6}\]
Substitute \(x\)\[y=\frac{4*3^2+12*3}{3^2-3+6}\]
\[y=\frac{36+36}{9-3+6}\]\[y=\frac{72}{12}\]\[y=6\]
Since \(y\) is a positive number in that equation, there is no hole at point (3,6)
Just keep trying for all possible combinations

Answer 4

I think that is how you are meant to solve it

Answer 5

so whats the answer

Answer 6

I have been told I can only tell you how to do it, not what the answer is. Sorry

Answer 7

http://www.purplemath.com/modules/grphrtnl4.htm

Answer 8

^^ That ^^

Answer 9

@ganeshie8

Answer 10

@Ahsome

Answer 11

i got -9

Answer 12

so b-3

Answer 13

?? ami correct

Answer 14

when is the denominator zero?
can you factor the denominator?

Answer 15

was i correct or not

Answer 16

and i did factor the bottom

Answer 17

when x= -9,
the denominator becomes
-9^2+-9-6
=-81-9-6
=-96
but you want to find =0


What did you get when you factored the denominator?

Answer 18

when i plugged in -3?

Answer 19

(-3)^2+-3-6
= 9-3-6
=0

Answer 20

so i was right the answer is B