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2 years ago

9

PLEASE HELPPPPP ASAP

1 answer:

saul85 [17]2 years ago

5 0

Answer:

it's the last one, the function is decreasing from x=-3 to x=-1

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<img src="https://tex.z-dn.net/?f=f%28x%29%20-%20%5Cfrac%7Bx%5E%7B2%7D-4%20%7D%7Bx%5E%7B4%7D%20%2Bx%5E%7B3%7D%20-4x%5E%7B2%7D-4%

Llana [10]

a) The given function is

$f(x)=\frac{x^2-4}{x^4+x^3-4x^2-4}$

The domain refers to all values of x for which the function is defined.

The function is defined for

$x^4+x^3-4x^2-4\ne0$

This implies that;

$x\ne -2.69,x\ne 1.83$

b) The vertical asymptotes are x-values that makes the function undefined.

To find the vertical asymptote, equate the denominator to zero and solve for x.

$x^4+x^3-4x^2-4=0$

This implies that;

$x= -2.69,x=1.83$

c) The roots are the x-intercepts of the graph.

To find the roots, we equate the function to zero and solve for x.

$\frac{x^2-4}{x^4+x^3-4x^2-4}=0$

$\Rightarrow x^2-4=0$

$x^2=4$

$x=\pm \sqrt{4}$

$x=\pm2$

The roots are $x=-2,x=2$

d) The y-intercept is where the graph touches the y-axis.

To find the y-inter, we substitute;

$x=0$ into the function

$f(0)=\frac{0^2-4}{0^4+0^3-4(0)^2-4}$

$f(0)=\frac{-4}{-4}=1$

e) to find the horizontal asypmtote, we take limit to infinity

$lim_{x\to \infty}\frac{x^2-4}{x^4+x^3-4x^2-4}=0$

The horizontal asymtote is $y=0$

f) The greatest common divisor of both the numerator and the denominator is 1.

There is no common factor of the numerator and the denominator which is at least a linear factor.

Therefore the function has no holes.

g) The given function is a proper rational function.

There is no oblique asymptote.

See attachment for graph.

6 0

2 years ago

1.) What is the equation of the path of firework #1? Write your equation in general form.

valina [46]

1. I'm assumig that the paths are perfect parabolas

this means that their general forms can be written in $y=ax^2+bx+c$

it's easier to find vertex form first then expand to get general form

vertex form is $y=a(x-h)^2+k$ where the vertex is (h,k) and a is a constant

firework #1

vertex is (10,50), so (h,k)=(10,50) and h=10, k=50

$h_1=a(t-10)^2+50$

to find the value of $a$ , subsitute another point

(0,0)

$0=a(0-10)^2+50)$

$0=100a+50$

$a=\frac{-1}{2}$

so the equation in vertex form is $h_1=\frac{-1}{2}(t-10)^2+50$

expand to get general form

$h_1=\frac{-1}{2}(t^2-20t+100)+50$

$h_1=\frac{-1}{2}t^2+10t-50+50$

$h_1=\frac{-1}{2}t^2+10t$

2.

same as last time

vertex is (10,72) so (h,k)=(10,72) so h=10 and k=72

equation is $h_2=a(t-10)^2+72$

find $a$

use another point

(0,22)

$22=a(0-10)^2+72$

$22=100a+72$

$-50=100a$

$a=\frac{-1}{2}$

so the equation in vertex form is $h_2=\frac{-1}{2}(t-10)^2+72$

3.

range is the numbers that h is allowed to be

think about what h represents. it represents the height of the rocket

from the graph, we can see that the lowest possible height is 0yd and the highest height is 50yd

so range is 0 to 50 or 0≤h≤50

domain is the numbers that t is allowed to be

think about what t represents. it represents how long the rocket has been flying

it will stop flying when it hits the ground or at t=20

it starts flying at t=0

so domain is from 0 to 20 or 0≤t≤20

3 0

2 years ago

Sora was given two data sets, one without an outlier and one with an outlier. Data without an outlier: 108, 113, 105, 118, 124,

gladu [14]

The outlier (61) is at the low end of the data set, but doesn't affect the mean by a lot, so ...

The mean is centered among the other numbers in both sets of data.

_____

The mean without the outlier is 114. With the outlier, it is 107.4. The lower quartile is 108, so the mean does get moved outside the "box" of the box-and-whisker plot of the data set without the outlier.

8 0

2 years ago

Read 2 more answers

Janiya just turned 18 years old. She spent 4/9 of her life in NY. How many years did Janiya live in NY? *

mr Goodwill [35]

4/9 x 2 = 8/18 so 8 yrs of her life

7 0

3 years ago

Which of the following rational numbers is equal to 3/5?

Vedmedyk [2.9K]

3/5 = 60%
20 ÷ 5 = 4
3 x 4 = 12

Answer is A) 12/20

4 0

3 years ago

Read 2 more answers