Question

The graphs of y=x-3 and y = 3x -4 intersect at

Answer 1

Answer:

Option 3) (0.38,-2.85) and (2.62,3.85)

Step-by-step explanation:

The correct question is

The graphs of y=x^2-3 and y=3x-4 intersect at approximately...

we have

$y=x^2-3$ ----> equation A

$y=3x-4$ ----> equation B

Solve the system by graphing

Remember that the solution of the system is the intersection point both graphs

Equate equation A and equation B

$x^2-3=3x-4$

$x^2-3x-3+4=0$

$x^2-3x+1=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$x^2-3x+1=0$

so

$a=1\\b=-3\\c=1$

substitute in the formula

$x=\frac{-(-3)\pm\sqrt{-3^{2}-4(1)(1)}} {2(1)}$

$x=\frac{3\pm\sqrt{5}} {2}$

so

$x_1=\frac{3+\sqrt{5}} {2}=2.62$

$x_2=\frac{3-\sqrt{5}} {2}=0.38$

Find the values of y (substitute the value of x in equation A or equation B)

For x=2.62 ----> $y=(2.62)^2-3=3.85$

For x=0.38 ----> $y=(0.38)^2-3=-2.85$

therefore

The intersection points are approximately  (0.38,-2.85) and (2.62,3.85)