bumblebee123
Junior Member
- Joined
- Jan 3, 2018
- Messages
- 200
I've been looking at this question for a while and I still can't get my head around it. Can anyone help to explain the answer?
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
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