Question

Explain how to sketch a graph of the function f(x) = x3 + 2x2 – 8x. Be sure to include end-behavior, zeroes, and intervals where the function is positive and negative.

Answer 1

Answer:

Step-by-step explanation:

First, for end behavior, the highest power of x is x^3 and it is positive. So towards infinity, the graph will be positive, and towards negative infinity the graph will be negative (because this is a cubic graph)

To find the zeros, you set the equation equal to 0 and solve for x

x^3+2x^2-8x=0

x(x^2+2x-8)=0

x(x+4)(x-2)=0

x=0   x=-4   x=2

So the zeros are at 0, -4, and 2. Therefore, you can plot the points (0,0), (-4,0) and (2,0)

And we can plug values into the original that are between each of the zeros to see which intervals are positive or negative.

Plugging in a -5 gets us -35

-1 gets us 9

1 gets us -5

3 gets us 21

So now you know end behavior, zeroes, and signs of intervals

Hope this helps

Answer 2

The zeroes of the function are -4, 0 and 2.

The intervals where the function is positive is $-4 < x < 2, \ \ x >2$.

The intervals where the function is negative is $x < -4$.

The given parameters:

• f(x) = x³ + 2x² - 8x

What is zeroes of a function?

The zeroes of a function is the possible values of the unknown that makes the entire function to be zero.

The zeroes of the cubic equation is calculated as follows;

f(x) = 0

x³ + 2x² - 8x

factorize as follows;

$x(x^2 + 2x -8) = 0\\\\x(x^2 + 4x - 2x -8) = 0\\\\x[x (x + 4 )-2(x + 4)]= 0\\\\x(x + 4)(x -2)=0\\\\x = 0, \ \ x = -4 \ \ x = 2$

The intervals where the function is positive and negative is determined as follows;

$x(x + 4) (x - 2)$

$when, \ x = -5, \ f(x) = -ve\\\\when, \ x = -4, \ f(x) =0\\\\when , \ x = -3, \ f(x) = +ve$

The intervals where the function is positive is determined as;

$-4 < x < 2, \ \ x >2$

The intervals where the function is negative is determined as;

$x < -4$

Learn more about graph of cubic equation here: https://brainly.com/question/8878887