Show whether or not y=x+3 is tangential to the curve y^2=x

The line y = x + 3 has slope 1, so we look for points on the curve where the tangent line, whose slope is dy/dx, is equal to 1.

y² = x

Take the derivative of both sides with respect to x, assuming y = y(x) :

2y dy/dx = 1

dy/dx = 1/(2y)

Solve for y when dy/dx = 1 :

1 = 1/(2y)

2y = 1

y = 1/2

When y = 1/2, we have x = y² = (1/2)² = 1/4. However, for the given line, when y = 1/2, we have x = y - 3 = 1/2 - 3 = -5/2.

This means the line y = x + 3 is not a tangent to the curve y² = x. In fact, the line never even touches y² = x :

x = y²   ⇒   y = y² + 3   ⇒   y² - y + 3 = 0

has no real solution for y.

hello from mrbilldoesmath!

answer:

25x^2 + 90x + 84

discussion:

mn(x) = m(n(x)) = m( 5x + 9) =

(5x + 9)^2 + 3 =>

25x^2 + 90x + 9^2 + 3 =

25x^2 + 90x + 84                 (as 9^2 = 9*9 = 81)

which is the second choice

regards,  

mrb

p.s.   i'll be on vacation from friday, dec 22 to jan 2, 2019. have a great new year!