Q2. - Algebra 2

Question

Identify the domain and range of the relation in the graph at the right. Predict the value of $\mathbf{y}$ when $\mathbf{x}=\mathbf{5}$.

Step-by-Step Solution

Answer

The domain and range of the given relation are $\left\{-1,1,2,4\right\}$ and $\left\{0,2,3\right\}$ respectively. The predicted value when $x=5$ is $y=4$.

Step 1 – State the concept

The x values of the ordered pairs belonging to a relation constitutes the domain of the relation and the y values of the ordered pairs belonging to a relation constitutes the range of the relation.

 

A line of fit and a prediction equation for a set of data can be found by selecting two points that appear to represent the data well.

 

The equation of a straight-line passing through the points $\left(h,k\right)$ and $\left(a,b\right)$ is given as $\frac{y-k}{x-h}=\frac{b-k}{a-h}$.

Step 2 – List the given data

A relation represented by the following graph:

So, the relation is the set of ordered pairs, $\left\{\left(-1,0\right),\left(1,2\right),\left(2,2\right),\left(4,3\right)\right\}$.

Step 3 – Determine the domain

The x values of the ordered pairs belonging to the relation $\left\{\left(-1,0\right),\left(1,2\right),\left(2,2\right),\left(4,3\right)\right\}$ are $-1,1,2,4$.

So, the domain of the relation is $\left\{-1,1,2,4\right\}$.

Step 4 – Determine the range

The x values of the ordered pairs belonging to the relation $\left\{\left(-1,0\right),\left(1,2\right),\left(2,2\right),\left(4,3\right)\right\}$ are $0,2,2,3$.

 

So, the range of the relation is $\left\{0,2,3\right\}$.

Step 5 – Prediction equation

Aside from the point $\left(1,2\right)$, the rest of the points of the relation appears to lie on a straight line. Choosing two of those points, $\left(-1,0\right)$ and $\left(2,2\right)$, the line of fit is given by the straight line passing through these points.

 

So, $\left(h,k\right)=\left(-1,0\right)$ and $\left(a,b\right)=\left(2,2\right)$.

 

Put $\left(h,k\right)=\left(-1,0\right)$ and $\left(a,b\right)=\left(2,2\right)$ in $\frac{y-k}{x-h}=\frac{b-k}{a-h}$ to get,

 

$\frac{y-0}{x-\left(-1\right)}=\frac{2-0}{2-\left(-1\right)}$

 

$\frac{y}{x+1}=\frac{2}{2+1}$  (Simplify)

 

$\frac{y}{x+1}=\frac{2}{3}$  (Simplify)

 

$\frac{y}{x+1}\left(x+1\right)=\frac{2}{3}\left(x+1\right)$  (Multiply both sides by $x+1$)

 

$y=\frac{2}{3}x+\frac{2}{3}$  (Simplify)

 

So, $y=\frac{2}{3}x+\frac{2}{3}$ is the prediction equation.

 

Put $x=5$ in $y=\frac{2}{3}x+\frac{2}{3}$ to get,

 

$\begin{array}{c}y=\frac{2}{3}\left(5\right)+\frac{2}{3}\\ =\frac{10}{3}+\frac{2}{3}\\ =\frac{12}{3}\\ =4\end{array}$

 

So, the predicted value when $x=5$ is $y=4$.