# Math Help - Leading ciefficient test

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 $-2x^9$