Question

please help me!!! ASAP!
1. what are the points of discontinuity?
y= x-8/x^2+6x-7

A. x=1, x=-7
B. x=-1, x=7
C. x=8
D. x=1,x=7

Answer 1

Factor out x^2+6x-7 first

Answer 2

A discontinuity will occur when the denominator = 0...

so what value(s) of 'x' will make your denominator = 0?

Answer 3

im so confused

Answer 4

Factor ou x^2+6x+7, and find what x's cannot equal 0.

Answer 5

*-7

Answer 6

how do you factor it like 6+7 or what ???

Answer 7

\[\large x^2 + 6x - 7\]

to factor this...we need to see what 2 numbers...multiply to make -7....but at the same time also add to make 6

can you think of 2 such numbers?

Answer 8

-1 times 7 ?

Answer 9

Perfect!...so we know that -1 times 7 = -7 and also -1 + 7 = 6

So what we do..is we write that like this

\[\large (x - 1)(x + 7)\]

Notice how we just wrote the -1 ad the 7 in there next to the 'x's

Now we just need to find when THAT = 0....well it looks like when x = 1 we will have that first parenthesis = 0......and since that is multiplied...everything times 0 = 0 so x = 1 will be 1 point...

and also the second parenthesis...when x = -7 we will have another 0....so there we have it

\[\large x =1 , x = -7\]

Answer 10

oh!! okay i get it thank you so much :)

Answer 11

Anytime hun :)