Where does the line 2x-y+2 = 0 meets the curve g(x) = x^2-3x-7 ?
We are given the equation of the line 2x-y+2 = 0
We will rewrite into the slope form.
==> y= 2x+2 ..........(1)
Also, given the equation of the curve g(x) = x^2 -3x -7
We need to find the intersection points.
Then we know that g(x) = y
==> x^2 -3x-7 = 2x+2
We will combine all terms on the left side.
==> x^2 -5x -9 = 0
Now we will use the formula to find the roots.
==. x1= ( 5 + sqrt(25+36) / 2 = (5+sqrt61)/2
==> x2= (5-sqrt61)/2
Then we have two intersection points :
==> y(x1) = 2(5+sqrt61)/2 + 2 = 7+sqrt61
==> y(x2) = 2(5-sqrt61)/2 + 2 = 7-sqrt61
Then the intersection points are ( (5+sqrt61)/2 , (7+sqrt61) ) and ( (5-sqrt61)/2 , (7-sqrt61) )