Q.66 - Intermediate Algebra

Question

For the function $g\left(x\right) = 16 - 36x^2$, find:

$ⓐ g(-1) ⓑ g\left(0\right) ⓒ g\left(2\right)$

Step-by-Step Solution

Answer

The value of $g(-1)$ for the given function is $-20$.

The value of $g\left(0\right)$ for the given function is $16$.

The value of $g\left(2\right)$ for the given function is $-128$.

Part (a) Step 1.

The given polynomial function is $g\left(x\right) = 16 - 36x^2$

The given value function is $g(-1), g\left(0\right), g\left(2\right)$

Part (a) Step 1. The given function is g ( x )   =   16   −   36 x 2 . The given value function is g ( - 1 ) Putting the x = - 1 in the given function.

The given function is $g\left(x\right) = 16 - 36x^2$

putting $x=-1$

$$\begin{gathered} g(-1) = 16 - 36{(-1)}^2 \\ g(-1) = 16 - 36\times 1 \\ g(-1) = 16 - 36 \\ g(-1) = -20 \end{gathered}$$

Part (b) Step 1. The given polynomial function is g ( x )   =   16   −   36 x 2

The given value function is $g\left(0\right)$

Part (b) Step 2. Putting the x = 0 in the given function.

The given function is $g\left(x\right) = 16 - 36x^2$

putting $x=0$

$$\begin{gathered} g\left(0\right) = 16 - 36{\left(0\right)}^2 \\ g\left(0\right) = 16 - 36\times 0 \\ g\left(0\right) = 16 - 0 \\ g\left(0\right) = 16 \end{gathered}$$

Part (c) Step 1. The given polynomial function is g ( x )   =   16   −   36 x 2

The given value function is $g\left(2\right)$

Part (c) Step 2. Putting the x = 2 in the given polynomial function.

The given function is $g\left(x\right) = 16 - 36x^2$

Putting $x=2$

$$\begin{gathered} g\left(2\right) = 16 - 36{\left(2\right)}^2 \\ g\left(2\right) = 16 - 36\times 4 \\ g\left(2\right) = 16 - 144 \\ g\left(2\right) = - 128 \end{gathered}$$