The parabolas defined by the equations $y=x^2+4x+6$ and $y=\frac{1}{2}x^2+x+6$
intersect at points $(a,b)$ and $(c,d)$, where $c\ge a$. What is $c-a$?
+128400 \( $y=x^2+4x+6$ and $y=\frac{1}{2}x^2+x+6$\) \($c\ge a$\)
Set these equal
x^2 + 4x + 6 = (1/2)x^2 + x + 6 ...... rearrange as
(1/2)x^2 + 3x = 0 multiply through by 2
x^2 + 6x = 0 factor
x ( x + 6) = 0 set both factors to 0 and solve for x and x = 0 or x = -6
And when x = 0, y = (0)^2 + 4(0) + 6 = 6
And when x = - 6, y = (-6)^2 + 4(-6) + 6 = 18
So...... the intersection points are (0, 6) and ( -6 , 18)
And c - a = 0 - (-6) = 6
Here's a graph : https://www.desmos.com/calculator/txramx7pv5
