Question

Let f(x) be a cubic polynomial function such that y=13(x-1)+4 is tangent to the graph of f(x)at (1, 4) and y=(x+1)+6 is tangent to the graph of f(x) at (-1, 6). Find the equation of f(x)

Answer 1

f(x) = ax^3+bx^2+cx+d
f'(x) = 3ax^2+2bx+c

f(1)=4
f(-1)=6
f'(1) = 13
f'(-1) = 1

So, rack 'em up and knock 'em down:

a+b+c+d = 4
-a+b-c+d = 6
3a+2b+c = 13
3a-2b+c = 1

(a,b,c,d) = (4,3,-5,2)

f(x) = 4x^3+3x^2-5x+2

See the graphs at

http://www.wolframalpha.com/input/?i=plot+y%3D4x%5E3%2B3x%5E2-5x%2B2,+y%3D13(x-1)%2B4,+y%3D(x%2B1)%2B6,+for+-2%3C%3Dx%3C%3D2