Answer :
Answer:
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.
Step-by-step explanation:
First problem: The hole is going to make both the bottom and the top zero.
So I start at the bottom first.
[tex]x^2-3x+2=0[/tex]
The left hand expression is factorable.
Since the coefficient of [tex]x^2[/tex] 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:
[tex](x-2)(x-1)=0[/tex].
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.
[tex]x^2-4x+4[/tex]
x=2
[tex]2^2-4(2)+4[/tex]
[tex]4-8+4[/tex]
[tex]-4+4[/tex]
[tex]0[/tex]
So we have a hole at x=2.
[tex]x^2-4x+4[/tex]
x=1
[tex]1^2-4(1)+4[/tex]
[tex]1-4+4[/tex]
[tex]-3+4[/tex]
[tex]1[/tex]
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 [tex]x^2[/tex] 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.
[tex]f(x)=\frac{(x-2)(x-2)}{(x-2)(x-1)}=\frac{x-2}{x-1}[/tex]
So now let's plug in 2 into the simplified version:
[tex]f(2)=\frac{2-2}{2-1}=\frac{0}{1}=0[/tex].
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 [tex]x^2[/tex] being 1.
I'm going to start with the factored forms this time:
The factored form of [tex]x^2-3x-4[/tex] is [tex](x-4)(x+1)[/tex] because -4(1)=-3 and -4+1=-3.
The factored form of [tex]x^2-5x+4[/tex] is [tex](x-4)(x-1)[/tex] because -4(-1)=4 and -4+(-1)=-5.
Look at [tex]\frac{(x-4)(x+1)}{(x-4)(x-1)}[/tex].
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.
[tex]\frac{x+1}{x-1}[/tex]
Plug in x=4 to find the corresponding y:
[tex]\frac{4+1}{4-1}{/tex]
[tex]\frac{5}{3}[/tex]
The hole is at (4, 5/3).
Third problem:
[tex]x^2-4x+4[/tex] has factored form [tex](x-2)(x-2)[/tex] because (-2)(-2)=4 and -2+(-2)=-4.
[tex]x^2-5x+4[/tex] has factored form [tex](x-4)(x-1)[/tex] 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.