100k views
2 votes
Please help this is pre calculus

Please help this is pre calculus-example-1
User Tuizi
by
7.9k points

1 Answer

4 votes

Explanation:

1. First things first, the binomial tells us the zeroes.


- (x - 2) {}^(2) = 0


(x - 2) {}^(2) = 0


x - 2 = 0


x = 2

This root has a multiplicity of 2, so this root will "bounces off the x axis"


(x + 4) = 0


x = - 4

Our x intercepts are

2 and -4

This root has a multiplicty of 1, so it will cross thorough the x axis.

Our leading degree is odd and we have a negative leading coefficient so

as x approaches ♾️, f(x) approaches negative ♾️

and as x approaches negative ♾️, f(x) approaches ♾️.

Our y intercept can be found by letting x=0,

so


- (0 - 2) {}^(2) (0 + 4) = - ( - 2) {}^(2) (0 + 4) = - (4)(4) = - 16

So our y intercept is -16.

For Rational Functions, if the numerator and denominator share the same factor they are considered a removable discontinuity or hole.

Set it equal to 0.


x + 5 = 0


x = - 5

So we have a hole at x=-5.

Next, simplify the fraction.


(x - 4)/(x - 6)

Now, plug in -5


( - 5 - 4)/( - 5 - 6) = ( - 9)/( - 11) = (9)/(11)

So we have a hole at (-5,9/11).

To find y intercept, let x=0,


(0 - 4)/(0 - 6) = (2)/(3)

To find Vertical Asymptote, set denominator equal to 0


x - 6 = 0


x = 6

Draw a vertical dotted line at x=6

To find HA, this is the only case since the numerator and denominator has the same degree.

So using the leading coefficients, divide the numerator leading coefficient/ denominator leading coefficient.

The leading coefficient for both is 1 so


ha = (1)/(1)

So our HA is y=1.

Draw a horizon tal dotted line at y=1

We have no slope asymptote.

The graph is above.

Part 3:

(3x-1)(2x+1)>0


3x - 1 = 0


3x = 1


x = (1)/(3)


2x + 1 = 0


2x = - 1


x = - (1)/(2)

We have three possible regions, when x is positive

  • When x is less than -1/2
  • When x lies between -1/2 and 1/3
  • When x is greater than 1/3

Pick a random number less than -1/2 and plug it in the inequality to see if it true.

Let use -1


(3( - 1) - 1)(2( - 1) + 1) = ( - 4)( - 1) = 4

Since 4>0, Whenever x is less than -1/2, the function is positive (greater than zero)

Let's try the next region.

Let use 0


(3(0) - 1)(2(0) + 1) = ( - 1)(1) = - 1

So this solution won't work

Let try the next region, let use 1.


(3(1) - 1)(2(1) + 1) = (2)(3) = 6

Since 6>0, Whenever x is greater than 1/3, the function is positive(greater than zero).

Our answer is

(-oo,-1/2) U (1/3,oo).

Part 4:

We know x cannot be -2 because


x + 2 = 0


x = - 2

So -2 can not be in our answer.


(x - 1)/(x + 2) \leqslant 2


x - 1 \leqslant 2(x + 2)


x - 1 \leqslant 2x + 4


- 5 \leqslant x

So


x \geqslant - 5

However x can not be -2. so we say

[-5,-2) U (-2,oo)

Please help this is pre calculus-example-1
Please help this is pre calculus-example-2
User Unos
by
7.9k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories