#### bumblebee123

##### Junior Member

- Joined
- Jan 3, 2018

- Messages
- 196

**question:**the curve y= ( (x^2)/3 ) - 4x+ 7 crosses the y-axis at the point P

Q is the point ( 6, -5 )

**a)**find the gradient of the line PQ

I managed to do this:

point P is when the curve crosses the y-axis, so this must be when x=0

to find the y-coordinate: y= ( 0/3) - (4 x 0 ) + 7

y = 7

to find the gradient on PQ: ( -5 - 7 ) / ( 6 - 0 ) = -2 which is correct

**b)**for the curve, find dy/dx

I found this to be dy/dx = ( 2x / 3 ) - 4

**c)**the gradient of the curve at point R is equal to the gradient of the line PQ. Show that R is vertically below the midpoint of the line PQ

Can anyone help to explain what to do? thanks