Q. 5 - Precalculus Enhanced with Graphing Utilities

Question

Graph each equation of the system. Then solve the system to find the points of intersection.

$\left\{\begin{array}{l}y=x^2+1\\ y=x+1\end{array}\right.$

Step-by-Step Solution

Answer

Solutions of system of equations$\left\{\begin{array}{l}y=x^2+1\\ y=x+1\end{array}\right.$ are $(0,1)\&(1,2)$and the graph is

Step 1. Given data

The given system of equations is

$\begin{array}{l}y=x^2+1 \cdots \left(i\right)\\ y=x+1 \cdots \left(ii\right)\end{array}$

Step 2. Graph of the system of equations

Plot the graph of $y=x^2+1 \& y=x+1$on same cartesian plane

Step 3. Solution of system

subtract equation ii from equation i

$$\begin{gathered} y-y=(x^2+1)-(x+1) \\ 0=x^2-x \\ x(x-1)=0 \end{gathered}$$

 By zero product property

$x=0$

or

$$\begin{gathered} x-1=0 \\ x=1 \end{gathered}$$

So solutions are $(0,1) \& (1,2)$

Step 4. Solution of system

Substitute $x=0$ in equation ii

$$\begin{gathered} y=x+1 \\ y=0+1 \\ y=1 \end{gathered}$$

Substitute $x=1$in equation ii

$$\begin{gathered} y=x+1 \\ y=1+1 \\ y=2 \end{gathered}$$

So solutions of the system are  $(0,1)\&(1,2)$

Step 5. Verification

Substitute$x=0 \&y=1$ in the system

$\left\{\begin{array}{l}1=0^2+1\\ 1=0+1\end{array}\right.\to \left\{\begin{array}{l}1=1\\ 1=1\end{array}\right.$

Substitute $x=1 \&y=2$ in system

$\left\{\begin{array}{l}2=1^2+1\\ 2=1+1\end{array}\right.\to \left\{\begin{array}{l}2=2\\ 2=2\end{array}\right.$

The system of equations is satisfied by both solutions 

so solutions $(0,1)\&(1,2)$ are correct

Step 6. Point of intersection

Locate the $(0,1)\&(1,2)$ for point of interception in the graph of a system of equations