How do you solve #y = 3x + 12# and #y = x^2 + 2x - 18#?

1 Answer
May 2, 2016

Answer:

The points of intersections are: #(6,30)# and #(-5,-3)#

Explanation:

We are looking for the intersection of a line and a parabola. In other words, we want the points where if we put the same #x# into both equations, we get the same #y#, therefore we should set the two #y#'s equal to each other to get the equation:

#3x+12 = x^2+2x-18#

which we can simplify to

#0=x^2-x-30#

Now we can use the quadratic equation to find the roots of our new quadratic where #a=1#, #b=-1#, and #c=-30#

#x=(1+-sqrt(1+120))/2 implies 6,-5#

To find the corresponding #y# values we can use either equation, but it's simpler to use the line:

#y=3x+12 implies 30,-3#

So the points of intersections are: #(6,30)# and #(-5,-3)#