yer but not on chip shop chips.

:p

Glad you spotted my deliberate mistake there 😉

lol, where u get all these smilies from?

because my chords prove it

At A Y=-X³ At B Y+d = -(X+c)³
y=-x^3
Y+d= -(X+c)³
Y+d= -X³-c³-3X²c-3Xc²
d= X³-c³-3X²c-3Xc²-Y (Y=-X³)
d= -c³-3X²c-3Xc²
d= -c³-3X²c-3Xc²
d= -c(c²+3X²+3Xc)

Gradient AB= d è -c(c²+3X²+3Xc)
c c
Gradient AB= -c²-3X²-3Xc
Gradient at a point when c=0 è 0-3X²-0= -3X²

y=x^2+2
At A Y=X² +2 At B Y+d = (X+c)²+2
Y+d= (X+c)²+2
Y+d= X²+c²+2Xc+2
d= X²+c²+2Xc+2-Y (Y=X²+2)
d= X²+c²+2Xc+2-X² -2
d= c²+2Xc
d= c(c+2X)

Gradient AB= d è c(c+2X)
c c
Gradient AB= c+2X
Gradient at a point when c=0 è 0+2X = 2X

I DID IT!!!!!!! WOO HOO

It can be done but i dont know how, here is an example of y=x^2:

Here are my workings out for the gradient of the line y=x²:
At A Y=X² At B Y+d = (X+c)²
Y+d= (X+c)²
Y+d= X²+c²+2Xc
d= X²+c²+2Xc-Y (Y=X²)
d= X²+c²+2Xc-X²
d= c²+2Xc
d= c(c+2X)

Gradient AB= d è c(c+2X)
c c
Gradient AB= c+2X
Gradient at a point when c=0 è 0+2X = 2X

You use y+c y+d to work it out i think, but i dont know how 🙁

You should end up with somthing like 3x which means if you choos 5 as a value for x the gradient at that point is 15.