Find the third degree polynomial whose graph is shown in the

figure.

choices are

f(x) = x^3 - x^2 - 2x + 2

f(x) = 1/4(x)^3 - 1/2(x)^2 - x+2

f(x) = 1/4(x)^3 -1/4(x)^2 + 2x + 2

f(x) = 1/2(x)^3 - 1/2(x)^2 - x + 2

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- Jun 16th 2007, 07:09 PMwvmcanelly@cableone.netCan any help with this one pleaseFind the third degree polynomial whose graph is shown in the

figure.

choices are

f(x) = x^3 - x^2 - 2x + 2

f(x) = 1/4(x)^3 - 1/2(x)^2 - x+2

f(x) = 1/4(x)^3 -1/4(x)^2 + 2x + 2

f(x) = 1/2(x)^3 - 1/2(x)^2 - x + 2 - Jun 16th 2007, 08:09 PMwvmcanelly@cableone.net
The answer I got was f(x) = x^3 - x^2 - 2x + 2

I think this is the answer can someone let me know if I'm correct. I'f not you don't have to give me the answer but an explanation on how to figure it out would be great.

Thanks - Jun 16th 2007, 08:16 PMJhevon
- Jun 16th 2007, 08:27 PMJhevon
the answer is the second one:

$\displaystyle f(x) = \frac {1}{4} x^3 - \frac {1}{2} x^2 - x + 2$

Why did i choose this one? it has all the desired features:

$\displaystyle f(2) = f(-2) = 0$ and $\displaystyle f(0) = 2$

The first graph i suggested had the correct x-intercepts and shape, but the y-intercepts were off. if you divide the original graph i suggested by 4, you would get the above graph

to refresh your memory, i suggested $\displaystyle f(x) = (x + 2)(x - 2)^2$, our answer has to be that graph, or some constant times that graph, as the above is. do you see why? - Jun 16th 2007, 08:48 PMSoroban
Hello, wvmcanelly!

Quote:

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

$\displaystyle \begin{array}{ccc}(a)\;f(x) & = &x^3 - x^2 - 2x + 2 \\

(b)\;f(x) &= &\frac{1}{4}x^3 - \frac{1}{2}x^2 - x + 2 \\

(c)\;f(x) & = & \frac{1}{4}x^3 -\frac{1}{4}x^2 + 2x + 2 \\

(d)\;f(x) & = & \frac{1}{2}x^3 - \frac{1}{2}x^2 - x + 2\end{array}$

The graph has an x-intercept at -2.

. . The function has a factor of: $\displaystyle (x + 2)$

The graph is tangent to the x-axis at 2.

. . The function has a factor of: $\displaystyle (x - 2)^2$

The cubic is of the form: .$\displaystyle f(x) \;=\;a(x+2)(x-2)^2$

Since $\displaystyle (0,2)$ is on the graph,

. . we have: .$\displaystyle a(0+2)(0-2)^2 \:=\:2\quad\Rightarrow\quad8a \,=\,2\quad\Rightarrow\quad a \,=\,\frac{1}{4}$

Therefore: .$\displaystyle f(x) \;=\;\frac{1}{4}(x+2)(x-2)^2 \;=\;\frac{1}{4}x^3 - \frac{1}{2}x^2 - x + 2$ . .Answer (b)

- Jun 16th 2007, 08:51 PMJhevon