Find all points, if any, where y - 4x = 12 intersects 2 - y = 2(x + 2)^2
Let's write the first as y = 4x + 12
And the second as y = 2 - 2(x + 2)^2
Set the y's equal and we have
2 - 2(x + 2)^2 = 4x + 12 simplify
2 - 2 [ x^2 + 4x + 4 ] = 4x + 12
2 - 2x^2 - 8x - 8 = 4x + 12
-2x^2 - 8x - 6 = 4x + 12 rearramge as
-2x^2 - 12x - 18 = 0 divide through by -2
x^2 + 6x + 9 = 0 factor
(x + 3)^2 = 0 take the square root
x + 3 = 0 subtract 3 from both sides
x = -3
And using y = 4x + 12
Then y = 4(-3) + 12 = 0
So....the solution is ( - 3, 0 )
Here's the graph that confirms this : https://www.desmos.com/calculator/6uplog6lqb