Find the equation of the line.

Question

Let f(x)=x^2 and g(x)=(x-74)^2+27. There is one line with positive slope that is tangent to both of the parabolas y=f(x) and y=g(x) simultaneously. 

Answer 1

A graphing calculator will show f(x) = x² "bottoming out" at (x,y) = (0, 0)

while g(x) = (x − 74)² + 27 is seen to reach its absolute minimum at

(x,y) = (74, 27).

[Note that the first derivative of g(x) is 2(x − 74) which equals zero at

x = 74; then g(74) gives 27 which also gives the lowest point of g(x)

as (x,y) = (74, 27).]

One would then simply draw a line through (0, 0) and (74, 27) for the

"one tangent line to both parabolas with positive slope" sought.

The equation of this line will be constructed as y = {(27 − 0)/(74 − 0)}x

or y = (27/74)x which has valid solutions at (0, 0) and (74, 27).