Question

ALGEBRA 2 HELP!!!
What is the discontinuity and zero of the function f(x) = the quantity of 3 x squared plus x minus 4, all over x minus 1?

Discontinuity at (-1, 1), zero at (four thirds, 0)

Discontinuity at (-1, 1), zero at (negative four thirds , 0)

Discontinuity at (1, 7), zero at (four thirds , 0)

Discontinuity at (1, 7), zero at (negtive four thirds, 0)

Answer 1

all of the above

Answer 2

??

Answer 3

@zepdrix

Answer 4

\[\Large\rm f(x)=\frac{3x^2+x-4}{x-1}\]This is our function?

Answer 5

yes

Answer 6

To find both `zeroes` and `discontinuities` we set our function equal to zero,\[\Large\rm 0=\frac{3x^2+x-4}{x-1}\]We'll look at the numerator and denominator separately.
The `numerator` will give us our zeroes.
The `denominator` will give us our discontinuities.

\[\Large\rm zeroes:\]\[\Large\rm 0=3x^2+x-4\]We'll have to throw this into the Quadratic Formula.\[\Large\rm x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]Remember how to do that?

Answer 7

not really

Answer 8

Our numerator is in the form:\[\Large\rm 0=ax^2+bx+c\]\[\Large\rm 0=3x^2+(1)x+(-4)\]

Answer 9

So we'll plug all those values in, and try to simplify.

Answer 10

\[\Large\rm x=\frac{-(1)\pm\sqrt{(1)^2-4(3)(-4)}}{2(3)}\]Understand where I got those numbers from? :o

Answer 11

yeah

Answer 12

So then uhhhh, what does that simplify down to?
Do it!! c:

Answer 13

i got (-1) i square root of -47 over 6 @zepdrix

Answer 14

im pretty sure thats wrong

Answer 15

Hmm under the root, see how we have subtraction AND a negative sign?

Answer 16

yeah

Answer 17

So that should change to positive, yes? :d

Answer 18

\[\Large\rm x=\frac{-1\pm\sqrt{(1)^2+4(3)(4)}}{6}\]

Answer 19

so -1 plus or minus 7 over 6

Answer 20

\[\Large\rm x=\frac{-1+7}{6}\]\[\Large\rm x=\frac{-1-7}{6}\]Ok good, so that gives us two values, yes?

Answer 21

so 1 and -8 over 6

Answer 22

Ok great. And -8 over 6 will simplify to -4 over 3.

Answer 23

We have to be careful though.
I probably should have mentioned this earlier.
It makes more sense to look for `discontinuities` FIRST.
If we ever come up with the same value for a `discontinuity` and a `zero`, then we do not include that value as one of our zeroes.

Answer 24

Let's jump to the denominator and see what's going on,\[\Large\rm 0=x-1\]

Answer 25

What value do you get for your discontinuity?

Answer 26

0

Answer 27

I set the denominator equal to zero, now we solve for x.

Answer 28

so we plug in what we just got?

Answer 29

No.
This is our denominator set equal to zero.\[\Large\rm 0=x-1\]Solve for x.
Isolate the x.
Get x alone.

Answer 30

1

Answer 31

Ok great, we find that our discontinuity is at x=1.

Answer 32

But wait that was one of our zeroes!
x=1,
x=-4/3

That means we cannot include x=1 as a zero of the function.

Answer 33

A discontinuity should NOT HAVE a y coordinate.
Whoever wrote this question is a silly billy.

We have a discontinuity at x=1.
And a zero located at (-4,3, 0).

Answer 34

so the answer is b!

Answer 35

Woops I made a typo, lemme do that again :(

Answer 36

And a zero located at (-4/3, 0).*

Answer 37

Ummmmm

Answer 38

Looks like it's D.
See how the x coordinate is 1 for the discontinuity?

Answer 39

yes thank you so much!

Answer 40

cool c: