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vovikov84 [41]
3 years ago
5

How do I find the "hole" in the graph of this rational function

Mathematics
1 answer:
aleksklad [387]3 years ago
4 0
Factor the denomenator
cancel out something
wait, what you canceld is now a hole

so
bottom factors to (x-3)(x-1)
so

(x-1)/[(x-3)(x-1)]
(x-1) factors out
set equal to 0 to find hole
x-1=0
x=1
at x=1 there is a zero
to find y value

get simplified equation
(x-1)/[(x-3)(x-1)] turned into x-3
for x=1
1-3=-2

the hole is at (1,-2)
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<h3>1 Answer: v = -130</h3>

=============================================================

Explanation:

Because we're given multiple choices to pick from, we can go through each to do trial and error.

For instance, if v = 52, then

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52/13 < -7

4 < -7

but that's a false statement since 4 is not smaller than -7. Use a number line to see this. You should have -7 to the left of 4, showing that -7 is the smaller value.

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Only v = -130 works since...

v/13 < -7

-130/13 < -7

-10 < -7

This is true because -10 is to the left of -7 on the number line.

-------------------------------------------------

Another approach:

We have some unknown number v, and we're dividing by 13 to get the expression v/13. To undo this, we multiply both sides by 13

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13*(v/13) < 13*(-7) .... multiply both sides by 13

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So any solution to the original inequality must be smaller than -91. This allows us to rule out v = 52, v = 26 and v = -39.

The value v = -130 is smaller than -91, so it satisfies v < -91. Therefore, v = -130 is a solution to the original inequality.

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