Question

For the function $f\left(x\right)=6x^2-3x-9$find

(a) $f\left(3\right)$

(b)$f\left(-2\right)$

(c)$f\left(0\right)$

Short answer

The required values are:

(a)$f\left(3\right)=36$

(b) $f(-2)=21$

(c)$f\left(0\right)=-9$

Step-by-step solution

Step 1. Given Information

The given function is$f\left(x\right)=6x^2-3x-9$.

Part (a) Step 1. Find the Value of Function at x=3

• Substitute $x=3$in $f\left(x\right)=6x^2-3x-9$and solve.

$\begin{array}{rcl}f\left(3\right)& =& 6{\left(3\right)}^2-3\left(3\right)-9\\ & =& 54-9-9\\ & =& 36\end{array}$

• So, the value of function at$x=3$is 36.

Part (b) Step 1. Given Information

The given function is $f\left(x\right)=6x^2-3x-9$.

Part(b) Step 2. Find the value of Function at 

• Substitute $x=-2$in $f\left(x\right)=6x^2-3x-9$and solve.

$\begin{array}{rcl}f\left(-2\right)& =& 6{(-2)}^2-3(-2)-9\\ & =& 24+6-9\\ & =& 21\\ & & \end{array}$
So, the value of function at $x=-2$is 21.

Part (c) Step 1. Given Information

The given function is$f\left(x\right)=6x^2-3x-9$.

Part (c) Step 2. Find the value of the Function at x=0

• Substitute $x=0$in $f\left(x\right)=6x^2-3x-9$and solve.

$\begin{array}{rcl}f\left(0\right)& =& 6{\left(0\right)}^2-3\left(0\right)-9\\ & =& -9\end{array}$

• So, the value of the function at $x=0$is $-9$.