Find all points, if any, where y - 4x = 12 intersects 2 - y = 2(x + 2)^2

travisio Feb 21, 2019

#1**+2 **

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

CPhill Feb 21, 2019