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Dmitry [639]
3 years ago
13

How do you graph x+2÷(x^2-3x-10) without a calculator?

Mathematics
1 answer:
ahrayia [7]3 years ago
5 0
First step is to simplify the equation into a form that we are familiar with graphing, in this case the simplest way i believe is to factorise the denominator. There are many methods of factorising quadratics but in this case my preferred method would be as follows:

x^2-3x-10, because it is a quadratic equation we know it can be factorised into the form (x+a)(x+b), we know that -10 is the product of a and b, therefore a and b can only be +/-1 and -/+10, or +/-5 and -/+2, the next step is to see which of these factors sum to equal -3, in this case it will be +2 and -5, therefore x^2-3x-10 can be factorised into (x+2)(x-5).

We have now simplified the function into f(x)= (x+2)/(x+2)(x-5), we can still simplify the function further though:

(x+2)/(x+2)(x-5), the numerator and denominator have a common factor therefore we can divide the numerator and the denominator by that factor, ((x+2)/(x+2))/((x+2)(x-5)/(x+2))= 1/(x-5) 

The function is now in it's simplest form, so we will now graph it. As a side note you can't perfectly "graph" any function by hand so instead we will sketch the function, to sketch a function you need to know the general shape of the function  and any x/y intercepts or asymptotes.        

Functions of the form a/(bx+c) are reciprocal functions, you need to be able to remember what reciprocal function looks like to be a able to sketch them by hand, they look something like this https://www.desmos.com/calculator/df2wiewhsz. To work out the x-intercept we will assign f(x) as 0, 0=1/(x-5), multiply both sides by (x-5), (x-5)0=1, 0=1, 0 doesn't equal 1 therefore there is no x-intercept but instead there is an asymptote at y=0. To work out the y-intercept we will assign x as 0, f(x)=1/(x-5), f(x)=1/-5, therefore the only y-intercept is at coordinates (0.-0.2). All reciprocal functions have 2 asymptotes, we have already found one of them, by making f(x)=0, to find the other asymptote we will make 1/(x-5)=∞, 1/0=∞, therefore x-5 has to equal 0, therefore the second asymptote will be x=5.

after putting all of this together the graph should look something like this https://www.desmos.com/calculator/bibeeu7xtv




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The standard linear equation is y=mx+b
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reciprocal (2 and -1/2)
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Mandarinka [93]
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Question is in the picture, please help me out!!
Grace [21]

The arithmetic sequences are as follow:

<h3>What is Arithmetic Sequence?</h3>

An arithmetic sequence in algebra is a sequence of numbers where the difference between every two consecutive terms is the same.

1) t(n) = 5n + 4

t(1) = 9, t(2) = 14, t(3) = 19

So,

9,14,19,...

d= 14-9 = 5

d= 19-14 =5

Hence, it is an AP

2) 1, 2, 4, 8 , 16

Hence, it is not an AP

3) 3, 6, 9 ,...

It is an AP

4)It is given that it is an AP

5) tn =  2*3^n

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So, 6, 18, 54,...

Hence, it is not an AP

6) 3 , 1, 1/3,...

It is not an AP

7) t(n+1)= 6*t(n)

t(1) = -1

t(2)=-6

t(3)= -36

Hence, it is not an AP

8) -3, 1, 5, 9

Hence, it is an AP.

9) 1, 4, 9,...

Hence, it not an AP

10) 2,1,0,1,2,...

It is not an AP

11) t(n)= -2n-5

t(1)= -7, t(2)= -9, t(3)= -11

Hence, it is an AP

12) tn= (1/2)^n

 t1= 1/2, t2= 1/4, t3= 1/8

It is not an AP

Learn more about AP here:

brainly.com/question/24873057

#SPJ1

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