Question

Please help this is pre calculus

Answer 1

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)