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.
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.
Hello, kwtolley!
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!