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## Thursday, March 19, 2009

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Welcome to the Jones College Prep Precalculus/IMP4 blog. This blog is a forum for students, teachers, and parents to discuss their math class at Jones College Prep.

## Thursday, March 19, 2009

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## 4 comments:

Sooo... just did the formative quiz that's on the website. I got that the third function is cubic. I know that the general equation for cubic functions is y=ax^3+bx^2+cx+d, but I don't quite know how to find a specific equation. I think I can find d (by plugging in the point 0,0).

Also, for the quadratic equations we used our knowledge that the second difference=2a. How can we find "a" in a cubic function without assuming that the third difference=3a? ...Or can we assume that?

Nadia,

We haven't officially covered how to get the specific equation for a cubic.

But you mentioned some good ideas. Using the point (0,0) will help you find d. And I like your prediction about the third difference, but I don't know if it works.

Can you make up a specific cubic function (maybe f(x)=x^3)and find the 3rd difference to see if there is a relationship between that constant number and "a"?

g(x) is a quadratic function.

To find the equation I used the general equation ax^2+bx+c and found c by pludding in "0" for x. C came out to be -2. Next, I used two points (-1,-1) and (1,-1) and plugged each set in to the general equation.

I ended up with

1)a-b-2=-1 and 2)a+b-2=-1

Then I isolated the "a" variable for 1 and got a=1+b

After that, I entered 1+b in to the second equation for "a" and ended up with

1+b+b-2=-1 or 1+2b-2=-1

Then I just solved for b and ended up with b=0.

Since b=0, I entered "0" in to the equation a=1+b and got a=1.

The equation = x^2-2

Alejandra,

Were you able to verify your equation, g(x)=x^2-2?

Could you use the same strategy to solve a cubic equation? What would you need to do differently?

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