Question

The solutions of the equation $x^2-2x-3=0$ are the _____-intercepts of the graph of$y=x^2-2x-3$.

Short answer

The solutions of the equation$x^2-2x-3=0$ are the x-intercepts of the graph of $y=x^2-2x-3$.

Step-by-step solution

Step 1. Apply the concept of intercepts.

The x-intercepts are the points on x-axis where the graph of the equation intersects the x-axis.

The y-intercepts are the points on y-axis where the graph of the equation intersects the y-axis.

The real solution of equation$y=f\left(x\right)$ are the points where the graph intersects the x-axis.

Step 2. Graph the equation.

Consider the equation $x^2-2x-3=0$.

Open the TI-83 calculator and follow the steps below.

Step 1: Press $\text{Y}=$key.

Step 2: Enter the function ${\text{Y}}_1=x^2-2x-3$.

Step 3: Press the WINDOW key.

Step 4: Make the following changes.

$\begin{array}{c}X\min =-7\\ X\max =10\\ Y\min =-10\\ Y\max =10\end{array}$

Step 5: Press the TRACE key.

The result obtained on screen is the graph of the equation $x^2-2x-3=0$.

The graph intersects thex-axis twice, so there are two real solutions of the equation.

Step 3. Find the x-intercept.

Once, the graph is obtained on screen, follow the steps below to compute the x-intercept.

Step 1: Press 2nd key and then TRACE key.

Step 2: Select the second option “zero”.

Step 3: To enter the left bound, move the curve to the left of the x-intercept and press ENTER key.

Step 4: To enter the right bound, move the curve to the left of the x-intercept and press ENTER key.

Step 5: For a guess value, press ENTER key.

The result obtained on screen is provided below.

Therefore, x-intercept is$-1$and one real solution of equation is $x=-1$.

Repeat the steps again.

Therefore, x-intercept is 3 and one real solution of equation is $x=3$.

There are two real solutions to equation as graph intersects the x-axis twice.

Thus, the solutions of the equation $x^2-2x-3=0$are the x-intercepts of the graph of $y=x^2-2x-3$.