OpenStudy (anonymous):

Determine all points of intersection parabola y=x^2+3x-4 and line y=5x+11

6 years ago
OpenStudy (jim_thompson5910):

\[y=x^2+3x-4 \] \[5x+11=x^2+3x-4 \] \[0=x^2+3x-4-5x-11 \] \[0=x^2-2x-15 \] \[0=(x-5)(x+3) \] \[x-5=0 \ \ \textrm{or} \ \ x+3=0\] \[x=5 \ \ \textrm{or} \ \ x=-3\] So the two solutions for x are x=5 or x=-3 When x=5, y = 5(5)+11 = 25+11 = 36 So when x=5, y = 36 giving us the ordered pair (5,36) When x=-3, y = 5(-3)+11 = -15+11 = -4 So when x=-3, y = -4 giving us the ordered pair (-3,-4) So there are two points of intersection and they are (5,36) and (-3,-4)

6 years ago
OpenStudy (anonymous):

Okay I think I get that, thanks again jim you're being a huge help

6 years ago
OpenStudy (jim_thompson5910):

glad to be of help

6 years ago