Question

Kelsey has a list of possible functions. Pick one of the g(x) functions below and then describe to Kelsey the key features of g(x), including the end behavior, y-intercept, and zeros..g(x)=(x+2)(x-1)(x-2).g(x)=(x+3)(x+2)(x-3).g(x)=(x+2)(x-2)(x-3).g(x)=(x+5)(x+2)(x-5).g(x)=(x+7)(x+1)(x-1).

Answer 1

Let's say the function is :

`g(x)=(x+2)(x-1)(x-2)`

To determine the end behavior, substitute a very large number to x, one for a negative value and one for positive.

So if we substitute a positive large number to x, the sign of the function will be :

`\begin{gathered} g(x)=(+)(+)(+) \\ g(x)=+ \end{gathered}`

The sign of g(x) will be positive.

And if we substitute a negative large number to x, the sign of the function will be :

`\begin{gathered} g(x)=(-)(-)(-) \\ g(x)=- \end{gathered}`

The sign of g(x) will be negative.

So we can conclude that as x goes to + infinity, g(x) goes to + infinity

and as x goes to - infinity, g(x) goes to - infinity

The y-intercept is the value of g(x) when x = 0

Substitute x = 0, then solve for g(x)

`\begin{gathered} g(x)=(0+2)(0-1)(0-2)_{} \\ g(x)=2(-1)(-2) \\ g(x)=-4 \end{gathered}`

Therefore, the y-intercept is at (0, 4)

For the zeros, the zeros are values of x when g(x) = 0

Set g(x) = 0, then solve for x :

`\begin{gathered} g(x)=(x+2)(x-1)(x-2) \\ 0=(x+2)(x-1)(x-2) \end{gathered}`

x + 2 = 0 => x = -2

x - 1 = 0 => x = 1

x - 2 = 0 => x = 2

The zeroes are :

(-2, 0), (1, 0) and (2, 0)