Find the third degree polynomial whose graph is shown in the figure. My answer is f(x)= 1/4x^3 - 1/4x^2 +2x+2, but I'm not sure. Thanks for checking my answer.

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- Jul 5th 2006, 09:08 AMkwtolleyThird degree
Find the third degree polynomial whose graph is shown in the figure. My answer is f(x)= 1/4x^3 - 1/4x^2 +2x+2, but I'm not sure. Thanks for checking my answer.

- Jul 5th 2006, 09:22 AMCaptainBlackQuote:

Originally Posted by**kwtolley**

RonL - Jul 5th 2006, 10:01 AMkwtolleycubic
The graph I put together is off some, as for being cubed or not thats why I'm asking. Some of my home work is hard, thats why I ask because I like to see how I messed up my answer, and checking the instructor also. Instructor/teachers can make mistakes just like all of us.

- Jul 5th 2006, 02:26 PMSoroban
Hello, kwtolley!

Quote:

Find the third degree polynomial whose graph is shown in the figure.

My answer is: $\displaystyle f(x) \:= \:\frac{1}{4}x^3 - \frac{1}{4}x^2 +2x+2$ , but I'm not sure.

We have a*big*problem . . .

A cubic function, $\displaystyle f(x) \;=\;ax^3 + bx^2 + cx + d$, has__four__constants to determine,

. . but we are given only__three__facts.

[1] The point $\displaystyle (0,2)$ is on the graph.

[2] The point $\displaystyle (-2,0)$ is on the graph.

[3] There is a critical point at $\displaystyle x = 2.$

Since $\displaystyle f(0) = 2:\;\;a\cdot0^3 + b\cdot0^2 + c\cdot0 + d\:=\:2\quad\Rightarrow\quad d = 2$

. . The function (so far) is: .$\displaystyle f(x)\;=\;ax^3 + bx^2 + cx + 2$

Since $\displaystyle f(-2) = 0:\;\;a(-2)^3 + b(-2)^2 + c(-2) + 2\;=\;0$

. . $\displaystyle -8a + 4b - 2c + 2\:=\:0\quad\Rightarrow\quad 8a - 4b + 2c\:=\:2$**[1]**

The derivative is: .$\displaystyle f'(x) \:=\:3ax^2 + 2bx + c$

Since $\displaystyle f'(2) = 0:\;\;3a\cdot2^2 + 2b\cdot2 + c\:=\:0\quad\Rightarrow\quad 12a + 4b + c \:=\:0$**[2]**

Add [1] and [2]: .$\displaystyle 20a + 3c\:=\:2\quad\Rightarrow\quad c\,=\,\frac{2-20a}{3}$

Substitute into [2]: .$\displaystyle 12a + 4b + \left(\frac{2 - 20a}{3}\right)\:=\:0\quad\Rightarrow\quad b = \frac{-8a - 1}{6}$

Therefore: .$\displaystyle \boxed{f(x) \;= \;ax^3 - \left(\frac{8a + 1}{6}\right)\!x^2 + \left(\frac{2 - 20a}{3}\right)\!x \,+ \,2\,}$

And, if there is no more information, that's the best we can do.

Edit: corrected that silly "3" . . . Thanks for the heads-up! - Jul 5th 2006, 03:48 PMQuickQuote:

Originally Posted by**Soroban**

- Jul 6th 2006, 08:03 AMkwtolleythanks
thanks for checking my answer I reworked it and got the right answer.