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How to find the equations of the lines that are tangent to two quadratic functions, f(x) = x² + 10 and g(x) = −(x − 8)² + 9, using the given equations of the slopes of the tangent lines. The steps to find the values of the variables a and c, and then calculates the equations of the tangent lines.
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This problem illustrates a method you can use to find lines that are tangent to two curves. Consider the quadratic functions
f (x) = x^2 + 10
and
g(x) = −(x − 8)^2 + 9
If we sketch their graphs on the same axes, we can see that there are two lines that are tangent to both curves:
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Let’s find the equations of those lines. Consider either line. Suppose it is tangent to y = f (x) at the point (a, b). Suppose it is tangent to y = g(x) at the point (c, d). This looks like we are introducing four variables: a, b, c, and d. But in fact there are only really two: since (a, b) is on the graph of y = f (x), we have
b = f (a)
Similarly,
d = g(c)
So we just need to find the values of a and c. If we could write two equations involving a and c, we’d be in business (we’d have ”two equations in two unknowns”, and that’s a good place to be algebraically). The way to get the two equations is to write the slope of the tangent line in several ways.
First, since the line passes through (a, b) and (c, d), we know the slope of the line is
m = d c −− ab
Also, since the line is tangent to y = f (x), we have
m = f ′(a)
and since the line is tangent to y = g(x), we also have
m = g′(c).
Putting these together, we get two equations:
d − b c − a =^ f^
′(a) (1)
and
f ′(a) = g′(c) (2)
Let’s see what each equation says. Equation (1) says
g(c) − f (a) c − a = 2a
so
g(c) − f (a) = 2a(c − a).
Using the definitions of g(x) and f (x), this equation becomes
−(c − 8)^2 + 9 − (a^2 + 10) = 2a(c − a). (3)
Let’s leave that equation alone for a moment and see what equation (2) says. We know
f ′(x) = 2x
and
g′(x) = −2(x − 8)