Polynomial functions: finding equation from graph, etc.  TOPIC_SOLVED

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Polynomial functions: finding equation from graph, etc.

Postby honest_denverco09 on Tue Apr 21, 2009 3:21 am

I have two questions :
1/ Please help me write a quadratic function whose graph has only one x-intercept, -4 and whose y-intercept, -8 in factored form. Thank a lot.
2/ If we have a graph of polynomial function, how can we write its equation ? Actually, if i have a graph of cubic or 4th-degree polynomial function, how can i know the sign of their leading coefficient, whether positive or negative ?
For EX : I have two following graphs :
One is quadratic function : Image
Two is 4th-degree function : Image

How can i write these equations ?
Thanks a lot!
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Postby stapel_eliz on Tue Apr 21, 2009 12:09 pm

honest_denverco09 wrote:1/ Please help me write a quadratic function whose graph has only one x-intercept, -4 and whose y-intercept, -8 in factored form.

I'm afraid I don't know what is meant by "the y-intercept 'in factored form'". The y-intercept is where the graph crosses the y-axis; "factoring" is not involved.

To learn how to find quadratics from their zeroes, try here. Thinking back to what you learned when you were factoring, solving, and graphing quadratics, you know that "one x-intercept" for a quadratic means one repeated root -- that is, there is one root that occurs twice -- and the graph touches the x-axis at the one zero (rather than passing through the axis).

honest_denverco09 wrote:2/ If we have a graph of polynomial function, how can we write its equation ? Actually, if i have a graph of cubic or 4th-degree polynomial function, how can i know the sign of their leading coefficient, whether positive or negative ?

To learn about polynomial behavior and the relationships between equations and graphs, try here.

honest_denverco09 wrote:For EX : I have two following graphs :
One is quadratic function : Image
Two is 4th-degree function : Image

How can i write these equations ?

Once you've studied the lesson on polynomial behavior, you will understand why it is obvious that neither of these pictures displays a quadratic or a quartic polynomial! :shock:

The first graph is a positive odd-degree polynomial, probably a cubic. The second graph is a negative odd-degree polynomial, possibly of degree five. :wink:
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Re: Polynomial functions: finding equation from graph, etc.

Postby honest_denverco09 on Wed Apr 22, 2009 2:11 am

Thank a lot!

I can not open the links that you sent to me, but from your answer, i guess so :

If the graph of quadratic or 4th-degree functions that have the start-point below the x-axis, their leading coefficients will be positive, and if they have the start-point above the x-axis, their leading coefficients will be negative.
For EX in these pictures, the first graph has the positive "a" and the second one has the negative "a".
I just guess so, is that true a little bit ? :D
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Postby stapel_eliz on Wed Apr 22, 2009 11:25 am

honest_denverco09 wrote:I can not open the links that you sent to me....

I'm sorry to hear there was some difficulty. The links are working now.

Please review the material, keeping in mind what you already know about graphing linear functions (degree one, and thus odd) and graphing quadratic functions (degree two, and thus even), as they are strongly indicative of what you will encounter for end-behavior in all even and odd polynomial functions. :wink:
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Re: Polynomial functions: finding equation from graph, etc.  TOPIC_SOLVED

Postby honest_denverco09 on Thu Apr 23, 2009 12:09 am

Thank a lot!
Last edited by honest_denverco09 on Thu Apr 23, 2009 11:20 pm, edited 1 time in total.
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Re: Polynomial functions: finding equation from graph, etc.

Postby honest_denverco09 on Thu Apr 23, 2009 11:20 pm

Here is the answer for my problem :

1/ We have x - intercept is only -4 , that means the vertex of this function's graph will be (-4, 0), and this x-intercept or this zero will be Double root.
As the theme, this polynomial function is a quadratic one, so it has two "square" (two zeros) or it is in form a(x - r1)(x - r2)
So, The equation will be y = a(x + 4)^2(x - r2)
We also have y-intercept is -8 , that means : The graph will be downward/the leading coefficient will be negative.
The point (0, -8) will be on the graph.
So far, we have : -8 = a(0 + 4)^2 and then solve for a, we'll get a = -2
Eventually, our equation is y = -2(x+4)^2

2/For this graph, i will do like this :
Image
When we have a graph of any polynomial function, first, we need to identify the sign of "a", leading coefficient whether positive or negative by looking at its graph. In the second picture, the "a" is negative.
Then, we need to realize what the degree of this function. In this case, the degree is 5.
Set up the parenthesis based on the degree and the graph. In this case, we will have y = -()()^3()
Seek for the x-intercepts to fill out the set above. In this case, we have x-intercepts are -5, -2, 1 .So, y = -(x+5)(x+2)^3(x-1)
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