The normal to the curve y = ax^1/2 + bx at the point where x = 1 has a gradient 1 and intercepts the y-axis (0,-14). Find a and b

Answer 1

y = a√x + bx
y' = a/(2√x) + b

Since the normal at x=1 has slope 1, the tangent has slope -1. So,

y'(1) = a/2 + b = -1
b = -(a+2)/2

y(1) = a - (a+2)/2 = (a-2)/2

The normal line through (1,(a-2)/2) is thus

y-(a-2)/2 = 1(x-1)
This passes through (0,-14), so
-14 - (a-2)/2 = 1(-1)
a = -24
So, b = 11

check:
y = -24/√x + 11x
y(1) = -13
the normal at (1,-13) is
y+13 = 1(x-1)
y = x-14
This passes through (0,-14)