# URGENT as level maths question help needed

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_martha

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#1

Can anyone tell me what to do to solve the question below?

A curve has equation y=g(x) where g(x) = kx³ + 15x² -39x + c

Given that the point P(2,10) lies on the curve and that the gradient of the curve at P is -3,

a) find the values of k and c

b) hence show that the curve has no stationary points

c) write g(x) in the form (x-4)h(x) where h(x) is an expression to be found

d) hence deduce the coordinates of the points of intersection of the curve with the equation y=g(0.2x) and the coordinate axis

If someone could even help with part a that would be really helpful as then I think I can do the rest. Thanks

A curve has equation y=g(x) where g(x) = kx³ + 15x² -39x + c

Given that the point P(2,10) lies on the curve and that the gradient of the curve at P is -3,

a) find the values of k and c

b) hence show that the curve has no stationary points

c) write g(x) in the form (x-4)h(x) where h(x) is an expression to be found

d) hence deduce the coordinates of the points of intersection of the curve with the equation y=g(0.2x) and the coordinate axis

If someone could even help with part a that would be really helpful as then I think I can do the rest. Thanks

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gdunne42

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#2

(Original post by

Can anyone tell me what to do to solve the question below?

A curve has equation y=g(x) where g(x) = kx³ + 15x² -39x + c

Given that the point P(2,10) lies on the curve and that the gradient of the curve at P is -3,

a) find the values of k and c

b) hence show that the curve has no stationary points

c) write g(x) in the form (x-4)h(x) where h(x) is an expression to be found

d) hence deduce the coordinates of the points of intersection of the curve with the equation y=g(0.2x) and the coordinate axis

If someone could even help with part a that would be really helpful as then I think I can do the rest. Thanks

**_martha**)Can anyone tell me what to do to solve the question below?

A curve has equation y=g(x) where g(x) = kx³ + 15x² -39x + c

Given that the point P(2,10) lies on the curve and that the gradient of the curve at P is -3,

a) find the values of k and c

b) hence show that the curve has no stationary points

c) write g(x) in the form (x-4)h(x) where h(x) is an expression to be found

d) hence deduce the coordinates of the points of intersection of the curve with the equation y=g(0.2x) and the coordinate axis

If someone could even help with part a that would be really helpful as then I think I can do the rest. Thanks

the gradient at P is -3, how do you find the gradient of a line from its equation? form another equation

Last edited by gdunne42; 1 month ago

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_martha

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#3

(Original post by

point P is on the line - sub in the values and form an an equation

the gradient at P is -3, how do you find the gradient of a line from its equation? form another equation

**gdunne42**)point P is on the line - sub in the values and form an an equation

the gradient at P is -3, how do you find the gradient of a line from its equation? form another equation

Last edited by _martha; 1 month ago

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gdunne42

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#4

(Original post by

How would I use this to find the values of c and k?

**_martha**)How would I use this to find the values of c and k?

what would you do to determine the gradient function of g(x)

(note that as c is a constant it disappears)

use the result to form another equation using (2,10) and -3

Last edited by gdunne42; 1 month ago

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_martha

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#5

(Original post by

what would you do to determine the gradient function of g(x)

(note that as c is a constant it disappears)

use the result to form another equation using (2,10) and -3

**gdunne42**)what would you do to determine the gradient function of g(x)

(note that as c is a constant it disappears)

use the result to form another equation using (2,10) and -3

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gdunne42

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#6

(Original post by

Do you mean that you would differentiate the original equation for g(x) or assume that the second differential is g''(x)=-3x+16 based on the information given about point P?

**_martha**)Do you mean that you would differentiate the original equation for g(x) or assume that the second differential is g''(x)=-3x+16 based on the information given about point P?

Last edited by gdunne42; 1 month ago

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#7

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I would differentiate g(x) to form another equation where you can substitute x=2 and g'(x)=-3 to solve k and hence solve c from your first result

**gdunne42**)I would differentiate g(x) to form another equation where you can substitute x=2 and g'(x)=-3 to solve k and hence solve c from your first result

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#8

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Thank you so much I've got it now!

**_martha**)Thank you so much I've got it now!

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