Q. 63 - Intermediate Algebra

Question

For the function $f\left(x\right)=8x^2-3x+2$, find:

$\left(a\right) f\left(5\right) \left(b\right) f(-2) \left(c\right) f\left(0\right)$

Step-by-Step Solution

Answer

The value of the function is 

$\begin{array}{l}\left(a\right)187\\ \left(b\right)40\\ \left(c\right)2\end{array}$

Step 1. Given information

$\begin{array}{l}\text{The given expressions is}\\ f\left(x\right)=8x^2-3x+2\\ \text{We have to find the value of the function }\\ \left(a\right)f\left(5\right)\text{ i.e at }x=5\\ \left(b\right)f(-2)\text{ i.e at }x=-2\\ \left(c\right)f\left(0\right)\text{ i.e at }x=0\end{array}$

Step 2. Solving for part (a)

$\begin{array}{l}\text{Substitute }x=5\text{ in the }f\left(x\right)\\ \Rightarrow f\left(5\right)=8{\left(5\right)}^2-3\left(5\right)+2\\ =8\left(25\right)-15+2\\ =200-15+2\\ =187\end{array}$

Step 3. Solving for part (b)

$\begin{array}{l}\text{Substitute }x=-2\text{ in the }f\left(x\right)\\ \Rightarrow f(-2)=8{\left(-2\right)}^2-3\left(-2\right)+2\\ =8\left(4\right)+6+2\\ =32+6+2\\ =40\end{array}$

Step 4. Solving for part (c)

$\begin{array}{l}\text{Substitute }x=0\text{ in the }f\left(x\right)\\ \Rightarrow f\left(0\right)=8{\left(0\right)}^2-3\left(0\right)+2\\ =8\left(0\right)-0+2\\ =0-0+2\\ =2\end{array}$