Question

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

Answer 1

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)`