# Homework Help: Sketch the graphs of the functions - Calculus question

1. May 15, 2013

### Mary4ever

1. The problem statement, all variables and given/known data
Sketch the graphs of the functions. Indicate intervals on which the function is increasing, decreasing, concave up, or concave down; indicate relative maximum points, relative minimum points, points of inflection, horizontal asympototes, vertical asymptotes, symmetry, and those intercepts that can be obtained conveniently:

2. Relevant equations
y=x^4 - 4x^3 - 20x^2 +150

3. The attempt at a solution
This is the solution I have:
This is a polynomial of degree 4, so there are no asymptotes.
The function has no symmetries.
dy/dx = 4x^3 - 12x^2 - 40x
d^2y/dx^2 = 12x^2 - 24x - 40

dy/dx = 4(x^3 - 3x^2 - 10x) = 4x(x-5)(x+2)

There are 0s at x = -2, 0, and 5
Thus, there are three critical points for a 4th degree polynomial.
As the function goes to infinity as x goes to plus or minus infinity, we know that
the function decreases on (-infinity, -2), increases on (-2, 0), decreases on (0,5), and increases on (5,infinity)
f(-2) = (-2)^4 -4*(-2)^3 - 20*(-2)^2 + 150 = 16 + 32 - 80 + 150 = 118
There is a local minimum at (-2, 118)
f(0) = 150
There is a local maximum at (0, 150)
f(5) = (5)^4 -4*(5)^3 - 20*(5)^2 + 150 = 625 - 500 - 500 + 150 = -225
There is a local minimum at (5, -225)

The y-intercept is 150

The 0s are approximately 2.5139 and 6.5252

Finally, as the second derivative is 12x^2 - 24x - 40 = 4(3x^2 - 6x - 10), the inflection points are
1 plus or minus sqrt(39)/3 is approximately -1.08166599946613 and 3.08166599946613

As the second derivative is quadratic with a positive leading coefficient, we then know that
the function is concave up on (-infinity, 1 - sqrt(39)/3) or (-infinity, -1.08166599946613 ) and
(1 + sqrt(39)/3,infinity) or (3.08166599946613, infinity) and
concave down on (1 - sqrt(39)/3,1 + sqrt(39)/3) or (-1.08166599946613,3.08166599946613)

2. May 15, 2013

### tiny-tim

Welcome to PF!

Hi Mary4ever! Welcome to PF!

(try using the X2 button just above the Reply box )

Yes, that all looks fine.

(I haven't checked the y coordinate calculations)

What is worrying you about that?

3. May 15, 2013

### Mary4ever

Could you please double-check it because I need to make sure everything is correct? Thank you!

4. May 16, 2013

### ArcanaNoir

This all looks good, based on inspection of the graph and a little help from WolframAlpha.