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aivan3 [116]
3 years ago
7

The graph showing here is the graph of which of the following rational functions?

Mathematics
1 answer:
MA_775_DIABLO [31]3 years ago
7 0

Answer: A

Step-by-step explanation:

The graph has a vertical asymptote at x = -2. In this situation, vertical asymptotes happen when the denominator equals zero, which leaves the function undefined. Therefore, find the answer choice whose denominator equals zero when x = -2.

A would be correct because when x = -2,

F(-2)=\frac{1}{-2+2}=\frac{1}{0}, which is undefined.

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What is the slope of a line that is parallel to the line y =<br> 3x + 2?<br> 4
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(X^2+x-6)(2x^2+4x) factor completely
MissTica

Answer:

2x\left(x-2\right)\left(x+3\right)\left(x+2\right)

Step-by-step explanation:

Lets go ahead and take a step by step approach to solving your problem.

First we must start by factoring x^2+x-6.

To do this, we must break the expression into groups! Although I will also provide a definition to help you understand! For ax^2+bx+c we need to find u, v > u\cdot \:v=a\cdot \:c and u+v=b which we will group into \left(ax^2+ux\right)+\left(vx+c\right).

The values we have are...a=1,\:b=1,\:c=-6 and u\cdot v=-6,\:u+v=1.

Now we need the factors of 6 which are 1, 2, 3 and 6. We also need the negative factors which you get simply by multiplying the positives by -1 or just reversing them.

Now we need to check for every two factors if u + v = 1.

\mathrm{Check}\:u=1,\:v=-6: \:u\cdot v=-6,\:u+v=-5 \Rightarrow  \mathrm{False}

\mathrm{Check}\:u=2,\:v=-3: \:u\cdot v=-6,\:u+v=-1 \Rightarrow  \mathrm{False}

\mathrm{Check}\:u=3,\:v=-2: \:u\cdot v=-6,\:u+v=1 \Rightarrow  \mathrm{True}

\mathrm{Check}\:u=6,\:v=-1: \:u\cdot v=-6,\:u+v=5 \Rightarrow  \mathrm{False}

Therefore u=3,\:v=-2.

Now we want to \mathrm{Group\:into\:}\left(ax^2+ux\right)+\left(vx+c\right). Which is \left(x^2-2x\right)+\left(3x-6\right).

Now we must factor x from x^2-2x.

Lets apply the exponent rule of a^{b+c}=a^ba^c which means x^2=xx.

Therefore we now have xx - 2x. Lets factor out the common term of x to get x(x - 2).

Now factor 3 out of 3x-6.

Rewrite 6 as 3 * 2. 3x-3\cdot \:2.

Now factor out the common term of 3 > 3\left(x-2\right).

x\left(x-2\right)+3\left(x-2\right) > Factor out the common term of x - 2 > \left(x-2\right)\left(x+3\right).

Now lets factor 2x^2+4x.

Apply the previous exponent rule of a^{b+c}=a^ba^c > x^2=xx > 2xx+4x.

Now rewrite 4 as 2 * 2 > 2xx+2\cdot \:2x.

Now factor out the common term of 2x to get 2x(x + 2).

Combine it all and we get 2x\left(x-2\right)\left(x+3\right)\left(x+2\right).

Hope this helps!

7 0
3 years ago
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