Question

Find the parabola with equation
y = ax^2 + bxwhose tangent line at (2, 12)
has the equation y = 14x − 16

Answer 1

dy/dx = 2ax + b

when x = 2, slope = 14
2a(2) + b = 14
4a + b = 14 or b = 14-4a

also (2,12) lies on the parabola
12 = 4a + 2b
12 = 4a + 2(14-4a)
12 = 4a + 28 - 8a
4a = 16
a=8
then b= 14-32 = -18

<b>so the parabola is y = 8x^2 -18x</b>

check:
dy/dx = 16x - 18
when x=2
dy/dx = 32-18 = 14
equation:
y-12 = 14(x-2)
y = 14x - 28 + 12 = 14x -16

all is good.