# Math Help - Can any help with this one please

1. ## Can any help with this one please

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

2. 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

3. Originally Posted by wvmcanelly@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
it seems my first post was incorrect (well, not completely). how did you get that answer. i do not think it is correct, since if we plug in -2 for x, we don't get 0. but clearly -2 is a root of the graph

4. Originally Posted by wvmcanelly@cableone.net
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
the answer is the second one:

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

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

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

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 $f(x) = (x + 2)(x - 2)^2$, our answer has to be that graph, or some constant times that graph, as the above is. do you see why?

5. Hello, wvmcanelly!

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

$\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: $(x + 2)$

The graph is tangent to the x-axis at 2.
. . The function has a factor of: $(x - 2)^2$

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

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

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

6. Originally Posted by Soroban
Hello, wvmcanelly!

The graph has an x-intercept at -2.
. . The function has a factor of: $(x + 2)$

The graph is tangent to the x-axis at 2.
. . The function has a factor of: $(x - 2)^2$

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

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

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

i had originally forgotten to make an arbitrary constant a part of the formula, but i figured out that it should be 1/4 by just looking at the graphs...but your way is much nicer