y = ax^3+bx^2+cx
y' = 3ax^2+2bx+c
y" = 6ax+2b
Since y"=0 at x = -1, -6a+2b=0
3a-b = 0
y'=4 at x = -1, so 3a-2b+c = 4
y=-5 at x = -1, so -a+b-c = -5
Now you have the three equations
3a-b = 0
3a-2b+c = 4
a-b+c = 5
y = x^3+3x^2+7x
The tangent line at (-1,-5) has slope 4, so it is y = 4x-1
See the graph at
http://www.wolframalpha.com/input/?i=plot+y%3Dx%5E3%2B3x%5E2%2B7x,+y%3D4x-1+where+-3+%3C%3D+x+%3C%3D+1
Determine a,b and c so that curve y= ax^3 + bx^2 + cx will have a slope of 4 at its point of inflection (-1,-5)
Help please!
1 answer