If the parabola $y_1 = x^2 + 2x + 7$ and the line $y_2 = 6x + b$ intersect at only one point, what is the value of $b$?
y = x^2 + 2x + 7
y = 6x + b
Set these equal
x^2 + 2x + 7 = 6x + b rearrange as
x^2 - 4x + (7 - b) = 0
If this only has one solution point....it must be that
(-4)^2 - 4(7 - b) = 0
16 - 28 + 4b = 0
-12 + 4b = 0
-12 = -4b divide both sides by -4
3 = b
This graph shows the intersection point of (2, 15)
https://www.desmos.com/calculator/mlvins5guo