Differentiation - Finding Tangent Equation for C Curve at Point P

[MathsTooHard]

Homework Statement

A curve has an equation .. y=x^3-5x^2+5x+2

a) find dy/dx in terms of x

The points P and Q lie on C. The gradient of C at both P and Q is 2. The x-coordinate of P is 3

b) Fine the x-coordinate of Q

c) Find and equation for the tangent to c at P, giving your answer in the form y=mx + c, where m and c are constants.

Homework Equations

The Attempt at a Solution

I'm not whether I got parts a or b right, however completely stuck on part c.

a) dy/dx = 3x^2 -10x +5

b)
y = ax+b

y = -1
a = 2
x = 3

-1 = (2*3) + b
-1 = 6 + b
-7 = b

y = 2x -7

So.

2x-7 = 3x^2 -10x +5
0 = 3x^2 -12x +12
0= x^2-4x+4
(x-2) (x-2)

So x =2?

Part c I have no idea where to start.

 

[Dick]

I think for b) they should have said "slope of the tangent" rather than "gradient". This quantity is dy/dx. You've already computed it in part a). Now they are saying dy/dx=2 at two points P and Q. Can you solve dy/dx=2 for x? They've already told you one of the roots is x=3 at P. There's another one and that must be Q. For c) the line goes through P and is tangent to C. So you know the x coordinate is 3 and the slope is 2. What else do you need to do to find the line equation?

 

[HallsofIvy]

Dick, apparently in England it is quite common to use the term "gradient" where we "Amurrincans" would say "derivative" or "slope of the tangent line".

a) dy/dx = 3x^2 -10x +5

Part (a) is correct.

b)
y = ax+b

y = -1
a = 2
x = 3

-1 = (2*3) + b
-1 = 6 + b
-7 = b

y = 2x -7

So.

2x-7 = 3x^2 -10x +5
0 = 3x^2 -12x +12
0= x^2-4x+4
(x-2) (x-2)

So x =2?

It's not at all clear to me what you are doing! Since you are told "The gradient of C at both P and Q is 2", you have immediately that at P and Q, 3x^2 -10x +5= 2 so 3x^2- 10x+ 3= 0. Knowing that the x coordinate of P is 3 helps factor that: x- 3 must be a factor so you get 3x^2- 10+ 3= (x- 3)(3x- 1)= 0. Yes, that checks. Since the x-coordinate of P is x, and Q is the other point at which the gradient (derivative) is 2, the other solution to that equation!

c) Find and equation for the tangent to c at P, giving your answer in the form y=mx + c, where m and c are constants.

Well, that's easy using the information you are given! You know that y= mx+ b must go through the point P which has x coordinate 3. What is the y coordinate of P? (Again, that's easy, you know that y= x^3- 5x^2+ 5x+ 2 goes through P.) You know that the slope of the tangent line, m, is the gradient at P and you are told that that is 2.

 

[Dick]

Dick, apparently in England, it is quite common to use the term "gradient" where we "Amurrincans" would say "derivative" or "slope of the tangent line".

Thanks, Halls. I'll try to adjust my colonial vocabulary.