Question

I need help with these removable discontinuities.
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Answer 1

First problem: a (2,0)

Second problem: b. none of these; the answer is (4, 5/3) which is not listed.

Third problem: b. none of the above; there are no holes period.

Explanation:

First problem: The hole is going to make both the bottom and the top zero.

So I start at the bottom first.

`x^2-3x+2=0`

The left hand expression is factorable.

Since the coefficient of
`x^2` is 1, you are looking for two numbers that multiply to be 2 and add to be -3.

Those numbers are -2 and -1 since (-2)(-1)=2 and -2+(-1)=-3.

The factored form of the equation is:

`(x-2)(x-1)=0`.

This means x-2=0 or x-1=0.

We have to solve both equations here.

x-2=0

Add 2 on both sides:

x=2

x-1=0

Add 1 on both sides:

x=1

Now to determine if x=2 or x=1 is a hole, we have to see if it makes the top 0.

If the top is zero when you replace in 2 for x, then x=2 is a hole.

If the top is zero when you replace in 1 for x, then x=1 is a hole.

Let's do that.

`x^2-4x+4`

x=2

`2^2-4(2)+4`

`4-8+4`

`-4+4`

`0`

So we have a hole at x=2.

`x^2-4x+4`

x=1

`1^2-4(1)+4`

`1-4+4`

`-3+4`

`1`

So x=1 is not a hole, it is a vertical asymptote. We know it is a vertical asymptote instead of a hole because the numerator wasn't 0 when we plugged in the x=1.

So anyways to find the point for which we have the hole, we will cancel out the factor that makes us have 0/0.

So let's factor the denominator now.

Since the coefficient of
`x^2` is 1, all we have to do is find two numbers that multiply to be 4 and add up to be -4.

Those numbers are -2 and -2 because -2(-2)=4 and -2+(-2)=-4.

`f(x)=((x-2)(x-2))/((x-2)(x-1))=(x-2)/(x-1)`

So now let's plug in 2 into the simplified version:

`f(2)=(2-2)/(2-1)=(0)/(1)=0`.

So the hole is at x=2 and the point for which the hole is at is (2,0).

a. (2,0)

Problem 2:

So these quadratics are the same kind of the ones before. They all have coefficient of
`x^2` being 1.

I'm going to start with the factored forms this time:

The factored form of
`x^2-3x-4` is
`(x-4)(x+1)` because -4(1)=-3 and -4+1=-3.

The factored form of
`x^2-5x+4` is
`(x-4)(x-1)` because -4(-1)=4 and -4+(-1)=-5.

Look at
`((x-4)(x+1))/((x-4)(x-1))`.

The hole is going to be when you have 0/0.

This happens at x=4 because x-4 is 0 when x=4.

The hole is at x=4.

Let's find the point now. It is (4,something).

So let's cancel out the (x-4)'s now.

`(x+1)/(x-1)`

Plug in x=4 to find the corresponding y:

`(4+1)/(4-1){/tex]</p><p>[tex](5)/(3)`

The hole is at (4, 5/3).

Third problem:

`x^2-4x+4` has factored form
`(x-2)(x-2)` because (-2)(-2)=4 and -2+(-2)=-4.

`x^2-5x+4` has factored form
`(x-4)(x-1)` because (-4)(-1)=4 and -4+(-1)=-5.

There are no common factors on top and bottom. You aren't going to have a hole. There is no value of x that gives you 0/0.