First graph your x and y intercept then pick any two points on the line.

that's the equation used to find your slope
Example:
(4,2) (8,6) let's say those are the points you pick from the line
so then you subtract 6-2=4 (6 is your y2 and 2 is your y1)
then subtracts 8-4=4 (8is your x2 and 4 is your x1)
then your left with

if you can reduce the reduce in this case you can which leaves you with

and that's how you find the slope
(x - 2)²(x+1)
(x-2)(x-2)
multiply the two brackets together
(x)(x)=x^2
(x)(-2)=-2x
(-2)(x)=-2x
(-2)(-2)=4
x^2-2x-2x+4
x^2-4x+4
(x^2-4x+4)(x+1)
multiply the brackets together
(x^2)(x)=x^3
(x^2)(1)=x^2
(-4x)(x)=-4x^2
(-4x)(1)=-4x
(4)(1)=4
x^3+x^2-4x^2-4x+4x+4
Answer:
x^3-3x^2+4
Let
denote the rocket's position, velocity, and acceleration vectors at time
.
We're given its initial position

and velocity

Immediately after launch, the rocket is subject to gravity, so its acceleration is

where
.
a. We can obtain the velocity and position vectors by respectively integrating the acceleration and velocity functions. By the fundamental theorem of calculus,


(the integral of 0 is a constant, but it ultimately doesn't matter in this case)

and



b. The rocket stays in the air for as long as it takes until
, where
is the
-component of the position vector.

The range of the rocket is the distance between the rocket's final position and the origin (0, 0, 0):

c. The rocket reaches its maximum height when its vertical velocity (the
-component) is 0, at which point we have


I hope this helps you
4,8 (x+4)=2,16
48 (x+4)=21,6
x+4=21,6/48
x+4= 0,45
x=0,45-4
x= -3,55