# Polynomial function

1. Apr 30, 2004

### mustang

Graph each function and find its real zero.
Problem 1.
f(x)=x^3-2x^2-5x+6
Can you show it being solved using the derivative. This is the way i teacher showed it to us but we didn't have another term like in this problem the number 6.

2. Apr 30, 2004

### turin

I think this is Newton's method. Do you remember your teacher mentioning Newton's method?

3. Apr 30, 2004

### faust9

What's the derivative of a constant? If you take the first derivative you'll get the slope which allows you to solve for the real zeros, so knowing the first derivative is the slope then the derivative a constant would be? Also, do you know the general shape of polynominials raised to an even power and those raised to an odd power? Just knowing what to expect for a $x^2,x^3,x^4,x^5$ can really be a boon to figuring out how to approach real zeros and how many zero's to expect.

4. Apr 30, 2004

### mustang

This Is what we learned

The problem our teacher gave us to help explain how to solve polynomial function:
y=x(x+2)(x-4)
x-intercept:1,-2,4

x(x+2)(x-4)
x(x^2-2x-8)
x^3-2x-8
3x^2-4x-8=0 <====The derivative

x=-b+/-sqrt(b^2-4ac) 4+/- sqrt[(-4)^2-4(3)(-8)]
-------------------- = ----------------------------= 4+/-sqrt(112)
2a 2(3) ------------
6

4+/-sqrt(112)
------------
6
this value was then plug back into x^3-2x-8 so to know the y-axis value of this x-coordnate.
After that i plot the x and y intercepts and the points found using the quardratic forumla to draw the graph.

5. May 1, 2004

### faust9

using tex will make your presentation much easier to read. Go through a couple of threads such as this: https://www.physicsforums.com/showthread.php?t=8997 for further information. You can click on the code $Click\ Here$ to see how to enter the correct statements.

You have x-intercept:1,-2,4 . It shoud be 0,-2,4 because to find the x-int we set the y value to zero and set each term x equal to that:

$y=x(x+2)(x-4)$ thus $0=x,\ \ 0=(x+2),\ \ 0=(x-4)$ then solve each term for x.

how did you get this? I hope this is a type-o because you shouls have $x^3-2x^2-8x$. You got the correct derivative so I'm "assuming" you had the correct polynomial otherwise your derivative is wrong for $x^3-2x-8$.

The general procedure is take the first derivative and solve for all values of x. Plug each x value back into the original equation to get the associated y values. Take the second derivative (if you've studied using this for graphing otherwise disregard) to find concavity and local min/max values.

What you entered above was correct except for the things I pointed out. Just a hint for graphing, learn the basic shapes for polynomials of different powers, log function shapes, the shapes of square and cube root functions, and the shape of exponential functions, oh and the trig functions. If you know what to expect (ie even power polynomials have both legs pointing in the same direction) then you can check your work to see if it makes sense.

So, do you know what the first derivative for [itex] f(x)=x^3-2x^2-5x+6[/tex] is?

6. May 1, 2004

### HallsofIvy

Staff Emeritus
The phrasing of the original post was misleading. Your teacher did not "use the derivative" to find the real zeros- since the example you give was already factored, those were obvious. You teacher apparently used the derivative to find the slopes, max, min, etc.

The function of the problem, f(x)=x^3-2x^2-5x+6= (x-1)(x-3)(x+2) so the real zeroes (plural- your phrase "find its real zero"-singular- was also misleading) are 1, 3, and -2. The graph of the function goes through (-2,0), (1,0),m and (3,0).

7. May 1, 2004

### Parth Dave

There is a process ive always been taught. It's fairly lengthy however, and some of it you don't need to do, but it is just so you can be absolutely positive.

1. Find the intercepts. label them on your graph.
2. Find the domain (and range if it's possible) of the function: it will let you know if you made any mistakes in the end.
2. Behaviour: how does the graph seem to behave as it approaches the infinities. (ie. the limit as x approaches positive infinite, and the limit as x approaches negative infinite)
3. Take the first derivative: using the first derivative if dy/dx > 0 than the function is increasing. If dy/dx < 0 than the function is decreasing. It is extremely helpful to know the general behaviour of the graph at all points. If dy/dx = 0 that indicates a local maximum or minimum in f(x). The general method is to determine first where dy/dx = 0 or undefined. At these points the function can change whether it increases or decreases. Set up a table with all the values of when dy/dx = 0 or undefined. Find the corresponding y-values for each of these points and label them on your graph (if possible). Than check the behaviour of your function in between these points (ie increasing or decreasing). (remember, dont forget the infinites, you need to know whats happening in between each infinite and the local max/mins. closest to them)
4. Take the second derivative: if d^2y/dx^2 > 0 than the function is concave up. If d^2y/dx^2 < 0 than the function is concave down. If d^2y/dx^2 = 0, than the function has a point of inflection (which means it changes whether it is concave up or down). Use the same process as in step 3. Set up the range of values and check where the function is concave up and down.

Using all the information you obtained, you should have a very good idea of what the graph will look like. You can follow this process to graph anything. But like faust9 said, it is a great idea to know some basic ones to help you. You should be able to graph things like y = x, y = x^2, y = x^2, y = sin(x), y = cos(x), y = log(x), y = ln(x) and understand how to transform these equations. (ie what would happen if you add a constant, or how to translate them up/down/left/right.)

8. May 6, 2004

### slyboy

$$\frac{d^2y}{dx^2} = 0$$ does not necessarily mean you have a point of inflection. It might be a point of inflection, but it also might be a maximum or a minimum. Think about the behaviour of $$y=x^4$$ at $$x=0$$ for example.

You have to check all the derivatives to determine whether it is a point of inflection or not. The first non-zero derivative will tell you whether it is a maximum or a minimum, but if all derivatives are zero then you have a point of inflection. You can use the Taylor expansion to prove this.

Last edited: May 6, 2004