Chapter 7.3, Problem 5QR

Solution

a.

To identify: The $\left(-1,6\right)$ points are on the graph of the function.

  $\begin{array}{l}f\left(x\right)=2x^2-3x+1\\ \text{a}.\left(-1,6\right)\end{array}$

Yes $\left(-1,6\right)$ is on the graph of the function.

Given information:

  $\begin{array}{l}f\left(x\right)=2x^2-3x+1\\ \text{a}.\left(-1,6\right)\end{array}$

Explanation:

Consider the given information.

To determine out if a point is on a function, you must plug in the point to see if the function work.

  $\begin{array}{c}f\left(x\right)=2x^2-3x+1\\ 6=2{\left(-1\right)}^2-3\left(-1\right)+1\\ 6=2+3+1\\ 6=6\end{array}$

So, $\left(-1,6\right)$ is on the graph of the function.

b.

To identify: The $\left(2,1\right)$ points are on the graph of the function.

  $\begin{array}{l}f\left(x\right)=2x^2-3x+1\\ \text{b}\text{.}\left(2,1\right)\end{array}$

No $\left(2,1\right)$ is on the graph of the function.

Given information:

  $\begin{array}{l}f\left(x\right)=2x^2-3x+1\\ \text{b}\text{.}\left(2,1\right)\end{array}$

Explanation:

Consider the given information.

Once again, plug the point into the function. Because the answer does not equal the $y$ ,the point is not a part of the function

  $\begin{array}{c}f\left(x\right)=2x^2-3x+1\\ 1=2{\left(2\right)}^2-3\left(2\right)+1\\ 1=8-6+1\\ 1\ne 3\end{array}$

So, $\left(2,1\right)$ is not on the graph of the function.