Question
The equation of a curve is y=px+ h/x^3 ,(x not equal to 0) where h and k are constants. The
gradient of the curve y at the point (2,22) is −60. Hence, find the turning points of the curve and determine the nature of the turning points.
gradient of the curve y at the point (2,22) is −60. Hence, find the turning points of the curve and determine the nature of the turning points.
Answers
You must have done some work on it -- too bad you didn't see fit to post what you tried ...
y = px + h/x^3
y' = p - 3h/x^4
so, now we know that
2p + h/8 = 22
p - 3h/16 = -60
Now just solve for p and h.
y = -27/4 x + 284/x^3
y' = -(27/4 + 852/x^4)
since y' is never zero, there are no turning points.
The graph looks something like a hyperbola, with a slope asymptote (y' approaches -27/4 as x gets large)
y = px + h/x^3
y' = p - 3h/x^4
so, now we know that
2p + h/8 = 22
p - 3h/16 = -60
Now just solve for p and h.
y = -27/4 x + 284/x^3
y' = -(27/4 + 852/x^4)
since y' is never zero, there are no turning points.
The graph looks something like a hyperbola, with a slope asymptote (y' approaches -27/4 as x gets large)
thankyou so much for the help, i really appreciate it. thankyou thankyouu
Related Questions
x=2sint
y=-3cost
0<[smaller and equal to] t <[larger and equal to] pie
find a cartesian equ...
1. Given the curve
a. Find an expression for the slope of the curve at any point (x, y) on the cu...
There is a curve with equation y=3/x, xis not equal 0. Sketch the curve with equation y=3/(x+2), xis...
The slope of a curve is equal to y divided by 4 more than x2 at any point (x, y) on the curve.
A. F...