[RESOLVED] Problems with Differentiation

1. A curve has equation y = x(x-a)(x+a), where a is a constant. Find the equations of the tangents to the graph at the points where is crosses the x-axis.

There are 3 equations in total, but I cant seem to get the 3rd equation.
This is how I get the first 2 equations:

y = x³-a²x
dy/dx = 3x²-a²

(x-intercept, y=0)
x³-a²x = 0
x² = a²
x = a, -x = -a

3x²-a² = 3a²-a² = 2a²

Therefore, equations of tangents:
y = 2a²x-2a³, y = 2a²x+2a³

The third equation: y = -a²x
I wonder how to get the third equation?


2. Find the coordinates of the point of intersection of the tangents to the graph of y = x² at the points at which it meets the line with equation y = x+2

As for this question, I have completely no idea on what to do...the answer is (1/2,-2)

1. x=a
x=-a
and x=0

Im working on part 2!

1. 0 = x(x - a)(x + a)

Solutions x = a, x = -a GOT THESE and x = 0 MISSED THIS ONE

dy/dx = 3x² - a² So at point (0, 0) gradient is -a²

Stick into y - y1 = m(x - x1) straight line with gradient m and point (x1, y1)

2. First find points where graphs cross by solving x² = x + 2 to find x coordinates and then adding 2 (y = x + 2) to find the y coordinates


y = x² so dy/dx= 2x

Stick your x coordinates in to find the gradients of the tangents at each point

Use y - y1 = m(x - x1) to find the equations of each of the tangents (there should be two)

Rearrange the equations of both the tangents to get y =

eg y = m1x + c1 and y = m2x + c2

Solve m1x + c1 = m2x + c2 to find the x coordinate and substitute the value into either of the tangent equations to find the y coordinate

Just about to post a solution to (2) but was beaten to it.

Hey you beat me to 1 because I was busy typing away!

LOL...Anyway, thank you soooo much to both of you~ :bigyello: