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

Question

jim_thompson5910 · Answer

$$y=x^2+3x-4 $$


$$5x+11=x^2+3x-4 $$


$$0=x^2+3x-4-5x-11 $$


$$0=x^2-2x-15 $$


$$0=(x-5)(x+3) $$


$$x-5=0 \ \ \textrm{or} \ \ x+3=0$$


$$x=5 \ \ \textrm{or} \ \ x=-3$$


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)

anonymous · Answer

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

jim_thompson5910 · Answer

glad to be of help