calculus
y=x^3=cx and y=ax^2+bx1 have a common tangent at (1,4). find a, b, andc
asked by
help needed

I will assume the first equation is
y = x^3 + cx (since the + and + are just a "shift" away)
since (1,4) is a common point
4 = 1  c
c = 3 , and
4 = a(1) b  1
ab = 3 (#1)
If they have a common tangent, then their slopes must be the same.
3x^2 + c = 2ax + b , but c = 3
3x^2 + 3 = 2ax + b
again (1,4) must satisfy.
3 + 3 = 2a + b
2a  b = 6 (#2)
#2  #1 > a = 3
back in #1
3  b = 3
b = 0
a= 3, b=0, and c=3posted by Reiny
Respond to this Question
Similar Questions

maths
if y variesdirectly partly as b andc if y=3,b=2,andc=2,andwhenny=2,b=3 andc=1/2 find k hence findc when y=4,b=1 
Calculus  Tangent Line
Hi, im having problems with the following problem. The main issue is actually starting the problem. Find the two points on the curve y = x^4  2x^2  x that have a common tangent line. First, find the derivative of y(x) so that 
calculus
find,if any, an equation for a common tangent line to these 2 curves:5x^2;and 5x^2x+6 
Calculus
I have a two part question that pertains to a curve (r(x)) and its tangent line at x=3. We are given that at x=3, r(x)=8. In order to find the slope of the tangent line, we are given another point (on the tangent line): (3.2, 
Calculus  Maths
I got a few questions. Hope ya'll can help out. 1) for F(X) = 6x  2x^2 Find the gradient of the chord joining the point where the X coorinates are 1 and (1+h) respectively. b) hence find the gradient at x=1 2) Find the 
Calculus Help Please!!!
 Find the slope m of the tangent to the curve y = 4 + 4x^2 − 2x^3 at the point where x = a.  Find equations of the tangent lines at the points (1,6) and (2,4). (1,6) Y(x)= (2,4) Y(x)= 
math
1. two circles. x^2+y^216x8y+35=0 and x^2+y^2=5 are given. a. write the equations in the standard form. b.prove that the circles touch each other. c.hence calculate the equation of their common tangent 2. find the equation of a. 
calculus
find the equation of a quadratic function whose graph is tangent at x=1 to the line whose slope8, tangent at x=2 to solve the line with slope4 and tangent to the line y=8 
Calculus AB
Could someone please help me with these tangent line problems? 1) Find the equation of the line tangent to the given curve at the indicated point: 3y^3 + 2x^2 = 5 at a point in the first quadrant where y=1. 2) Show that there is 
Calculus Tangent lines
find the equation of a tangent line of x^(1/4) + y^(1/4) = 4 at the point (16,16)