Q. 23 - Precalculus Enhanced with Graphing Utilities

Question

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

$\left\{\begin{array}{l}y = x^2 - 4 \\ y = 6x - 13\end{array}\right.$

Step-by-Step Solution

Answer

the point of intersection is $(3,5)$.

Step 1. Given

$\left\{\begin{array}{l}y = x^2 - 4 \\ y = 6x - 13\end{array}\right.$

Step 2. Graphing each equation

$y = x^2 - 4$

$y = 6x - 13$

Step 3. Calculation of intersection point

$$\begin{gathered} Consider, \\ y = x^2 - 4 .....\left(1\right) \\ y = 6x - 13 .....\left(2\right) \\ Now substituting the value of y from eq.1 into eq.2 \\ we get \\ {x}^2-4=6x-13 \\ {x}^2-6x+9=0 \\ {(x-3)}^2=0x=3 \\ Now substituting value of x in eq. 1 we get \\ y={\left(3\right)}^2-4 \\ y=5 \end{gathered}$$

Step 4. Check the solution

$$\begin{gathered} Substitute the obtained values of x and y in both the equations \\ y = 6x - 13 \\ 5=6\left(3\right)-13 \\ 5=5 \\ also \\ y=x^2-4 \\ 5={\left(3\right)}^2-4 \\ 5=5 \\ holds true. \end{gathered}$$