How do you solve y=3x+12 and y=x2+2x−18?

1 Answer
David B.
May 2, 2016

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=x2+2x−18

which we can simplify to

0=x2−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±√1+1202⇒6,−5

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

y=3x+12⇒30,−3

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

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