21 POINTS There’s also things on my page worth lots points points as well! So if you could help with those that would be awesome!!

Second one

third one

fifth one

last one

Step-by-step explanation:

Based on the x intercepts we can write

a(x+1)(x-5)

where a is some constant

solve for a by plugging in some coordinates

let's plug in (1,-8)

-8=a(1+1)(1-5)

-8= -8a

a=1

therefore the quadratic is (x+1)(x-5)

expand this

x²-4x-5

To find out where something is decreasing/increasing it's easiest to take the first derivative

x²-4x-5= 2x-4

set this equal to 0

2x-4=0

2x=4

x=2

Which means that our two intervals are

(-∞,2)U(2,∞)

plug in any values in the intervals to see whether or not the function is increasing/decreation

Let's plug in 0 for (-∞,2)

2(0)-4

-4

It's negative so it's decreasing on this interval

DO the same thing with the other one

let's plug in x=3

2(3)-4

2

this is positive so it's increasing on this interval

Go back to the critical value of x=2 and plug this into the equation to find the max/min

(2+1)(x-5)= -9

This is a minimum because the cofeccient for the degree is positive

For the limit one notice that we have a positive, even degree which means the end behavior is positive, positive

what this means is as x approaches negative infinity it's infinity

and as x approaches infinity it's infinity

answer: the answer is one

step-by-step explanation:

pemdas, in this situation there is no parenthesis or exponents so you go to multiplication, 1x1=1, then division 1/1= 1 then addition 1+1=2 then subtraction 2-1=1