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 PM #1

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- Jun 16th 2007, 08:09 PM #2

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- Jun 16th 2007, 08:16 PM #3

- Jun 16th 2007, 08:27 PM #4
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 PM #5

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Hello, wvmcanelly!

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 PM #6