Use the leading coefficient test to indicate the end behavior of the graph of the polynomial function. Indicate the zero of the function.

f(x)=-2x^3(x+5)^4(x-3)^2

I need help with this problem. I don't even know where to begin.

2. Ok

Originally Posted by theevilp0ptart
Use the leading coefficient test to indicate the end behavior of the graph of the polynomial function. Indicate the zero of the function.

f(x)=-2x^3(x+5)^4(x-3)^2

I need help with this problem. I don't even know where to begin.
Is this supposed to be $f(x)=-2x^3\cdot(x+5)^4\cdot(x-3)^2$?

3. Originally Posted by Mathstud28
Is this supposed to be $f(x)=-2x^3\cdot(x+5)^4\cdot(x-3)^2$?
Yes.

4. Originally Posted by theevilp0ptart
Use the leading coefficient test to indicate the end behavior of the graph of the polynomial function. Indicate the zero of the function.

f(x)=-2x^3(x+5)^4(x-3)^2

I need help with this problem. I don't even know where to begin.
So the degree of f(x) is p and the lead terms will be of the form

$-2x^9$ (why?)

as $x \to -\infty \mbox{ f(x) } \to \infty$

as $x \to \infty \mbox{ f(x) } \to -\infty$

again why?

Since it is already factored the zero's are $x=0,-5,3$

Good luck.

5. Originally Posted by TheEmptySet
So the degree of f(x) is p and the lead terms will be of the form

$-2x^9$ (why?)

as $x \to -\infty \mbox{ f(x) } \to \infty$

as $x \to \infty \mbox{ f(x) } \to -\infty$

again why?

Since it is already factored the zero's are $x=0,-5,3$

Good luck.
I do not understand why it would be $-2x^9$

6. Originally Posted by theevilp0ptart
I do not understand why it would be $-2x^9$
What would happen if we multiplied out the whole thing?

Lets look at one part

$(x+5)^4=(x+5)(x+5)(x+5)(x+5)$

We distributedthe whole thing out (we don't need to) to find the highest power of x .

Well to get that terms we would multiply x by x by x by x $=x^4$

$
f(x)=-2x^3 \cdot \underbrace{(x+5)^4}_{x^4} \cdot \underbrace{(x-3)^2}_{x^2}
$

so our lead term will be $-2x^9$