Answer:
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
