# 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...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

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) )

Approved by eNotes Editorial Team