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
[tex]y=x^2-3[/tex] ----> equation A
[tex]y=3x-4[/tex] ----> 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
[tex]x^2-3=3x-4[/tex]
[tex]x^2-3x-3+4=0[/tex]
[tex]x^2-3x+1=0[/tex]
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]x^2-3x+1=0[/tex]
so
[tex]a=1\\b=-3\\c=1[/tex]
substitute in the formula
[tex]x=\frac{-(-3)\pm\sqrt{-3^{2}-4(1)(1)}} {2(1)}[/tex]
[tex]x=\frac{3\pm\sqrt{5}} {2}[/tex]
so
[tex]x_1=\frac{3+\sqrt{5}} {2}=2.62[/tex]
[tex]x_2=\frac{3-\sqrt{5}} {2}=0.38[/tex]
Find the values of y (substitute the value of x in equation A or equation B)
For x=2.62 ----> [tex]y=(2.62)^2-3=3.85[/tex]
For x=0.38 ----> [tex]y=(0.38)^2-3=-2.85[/tex]
therefore
The intersection points are approximately (0.38,-2.85) and (2.62,3.85)
