find the equation of a line

Question

Let f(x)=x^2 and g(x)=(x-1)^2+7. There is one line with positive slope that is tangent to both of the parabolas y=f(x) and y=g(x) simultaneously.

 

y=?

Answer 1

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Answer 2

Since the line is a tangent to both the parabolas simultaneously, first find if the two parabolas intersect. 

In this case they do at point (4, 16). This point of intersection can be found by equating x^2 = (x-1)^2 + 7.

Now the first derivative of any curve gives the slope of the tangent to the curve.

Thus, taking the first derivative of any of the above curves and substituting the value of x = 4, gives the slope of the tangent at point x = 4. 

Thus, 

f`(x) = 2x

At x = 4, the slope of the tangent is 8.

 

Now the general equation of any line is given by y = mx + c where m = slope and c = y-intercept.

In this case, we know the slope is 8 and we also know one point i.e. (4, 16) lies on the line.

Therefore, we can substitute all the above quantities to find the value of c. 

Knowing the value of m and c thus yields the equation of the line as 

y = 8x - 16.