Expert: Scotto - 12/31/2008

Question
Please,
1. The tangent point between g(x)=ax^2+bx+c and f(x)=mx+n is x1 resp y1. What's the slope of f(x)=mx+n? (m,n=?)
and vice versa:
2. If I have f(x)= mx+n what is a,b,c=? as long as g(x)=ax^2+bx+c is tangent in x1,y1
Thanks a lot

Answer
1. The slope of the tangent line is the same as g'(x).

This would mean that m would have to be 2a(x0) + b where x0 is the point where the line f(x) crosses g(x).

To find n, note that f(x0) = g(x0) and solve for n in terms of a, b, and c.

2. If f(x) = x-1, then the slope is 1.  Lets suppose it touches a parabola at x=1, so then we can say that the value f(1)=0.  We need a parabola that has slope one and g(1)=0.  

Since we know that g(x) = ax² + bx + c, it is known that
g(1) = 0 and g'(1) = 1.

The value g(1) = a + b + c = 0 and the slope g'(1) = 2a + b = 1.

Case 1
--------
If a=1, then from the value of g'(1), 2a + b = 1, b=-1.

Since a=1, b=-1, then from g(1) = a + b + c = 0, then c=0.

The equation of g(x) is then g(x) = x² - x.

Checking it out, g(1) = 1 - 1 = 0 and g'(1) = 2 - 1 = 1.

Case 2
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If a=2, then from the value of g(1), 2a + b = 1, so b=-3.

Since a=2 and b=-3, then from g(1) = a + b + c = 0, c = 1.

Another equation with this property would be g(x) = 2x² - 3x + 1.

Checking it out, g(1) = 2 - 3 + 1 = 0 and g'(1) = 2(2) - 3 = 1,
so this equation also holds.

From here, it can be seen that one of the values of a, b, or c needs to be known.  There is an infinite number of parabolas that are tangent to the line.