Question

f(x)=x^3-x^2-36x+36/x^3-17x^2+90x-144,
f(x) has one hole point at(__,__)

Answer 1

someone really wants you to learn to factor

Answer 2

well i know how to factor that out but i dont know what the hole point is

Answer 3

the hole appears when the top and bottom have a like factor

Answer 4

determine the common roots and youll see your "x" value; simplify and input the x again to get the y

Answer 5

never mind this one doesnt seem like it factors nicely

Answer 6

it will factor, find 1 rational factor and then use synthetic division to get the rest

Answer 7

for example:
\[\frac{(x-1)(x+2)}{(x+2)(x-3)}\]
x+2 is common and has a root at -2

\[\frac{(-2-1)\cancel{(x+2)}}{\cancel{(x+2)}(-2-3)}=\frac{3}{5} \]

Answer 8

even easier way is to use grouping method on first fraction

Answer 9

numeratpr(

Answer 10

i tried factoring but cant do i have to use the quadratic formula or am i just making it harder than what it is

Answer 11

x^3-x^2-36x+36
x^2(x-1)-36(x-1)
(x-1)(x^2-36)
(x-1)(x-6)(x+6) <===numerator

Answer 12

Quadratic formula is for quadratics, not cubics

Answer 13

lol "someone really wants you to learn to factor"

And this is why high school math before AP sucked.

Answer 14

For the denominator, i'm going to find a zero on a calculator and use that to try synthetic division

Answer 15

ok thanks cause i get for the denominator x^2(x-17)18(5x-8) and that cant be right

Answer 16

x^3-17x^2+90x-144,
3| 1 -17 90 -144
3 -42 144
---------------------------
1 -14 48 | 0

x^2-14x+48=0
(x-6)(x-8)=0
denominator is (x-3)(x-6)(x-8)

So in factored form numerator/denominator:
\[\frac{(x-1)(x-6)(x+6)}{(x-3)(x-6)(x-8)}=\frac{(x-1)(x+6)}{(x-3)(x-8)}\]

Answer 17

ok thanks for doing that but how do i find the hole

Answer 18

to quote amistre64: "the hole appears when the top and bottom have a like factor"

There is one set of like factors in the numerator and denominator: x - 6
So, there would be a hole at x - 6 = 0, or x = 6

Answer 19

i've actually got 2 holes

Answer 20

one at x=3, x=8

Answer 21

actually, x=3,x=8 are vertical asymptotes, not holes

Answer 22

x=6 would be the hole

Answer 23

so its 6,0 and you got the six becasue thats what was common in both of them and you canceled it

Answer 24

f(x) is undefined at x=3,6,8. But because (x-6) is also a factor in the numerator, it is only a hole in the graph, not a vertical asymptote

Answer 25

ok im just trying to figure out how to write it cause it gives me 2 spaces to enter a number and the first space is 6 but i cant fiugure out what goes in the next space

Answer 26

They want a y-value then?

Answer 27

\[\frac{(x-1)(x+6)}{(x-3)(x-8)}\], plug 6 into this expression i posted above

Answer 28

You get the point (6,-10)

Answer 29

ok got it thanks so much