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2 answers:
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There are two ways to do this but the way I prefer is to make one of the equations in terms of one variable and then 'plug this in' to the second equation. I will demonstrate
Look at equation 1,
this can quite easily be manipulated to show 
.
Then because there is a y in the second equation (and both equations are simultaneous) we can 'plug in' our new equation where y is in the second one
which can then be solved for x since there is only one variable 
and then with our x solution we can work out our y solution by using the equation we manipulated 
.
So the solution to these equations is x=-2 when y=6
Look at equation 1,
Then because there is a y in the second equation (and both equations are simultaneous) we can 'plug in' our new equation where y is in the second one
So the solution to these equations is x=-2 when y=6
Answer:
Intercepts:
x = 0, y = 0
x = 1.77, y = 0
x = 2.51, y = 0
Critical points:
x = 1.25, y = 4
x = 2.17 , y = -4
x = 2.8, y = 4
Inflection points:
x = 0.81, y = 2.44
x = 1.81, y = -0.54
x = 2.52, y = 0.27
Step-by-step explanation:
We can find the intercept by setting f(x) = 0
where n = 0, 1, 2,3, 4, 5,...
Since we are restricting x between 0 and 3 we can stop at n = 2
So the function f(x) intercepts at y = 0 and x:
x = 0
x = 1.77
x = 2.51
The critical points occur at the first derivative = 0
or
where n = 0, 1, 2, 3
Since we are restricting x between 0 and 3 we can stop at n = 2
So our critical points are at
x = 1.25,
x = 2.17
,
x = 2.8,
For the inflection point, we can take the 2nd derivative and set it to 0
We can solve this numerically to get the inflection points are at
x = 0.81,
x = 1.81,
x = 2.52,