Co-ordinates of point - Help Please... (1 Viewer)

[Smile12345]

Can anyone please help me with this question...

Find the coordinates of the point at which y=x^3 + 1 has a tangent with gradient of 3...

Thanks in advance...

 

[albertcamus]

dy/dx = 3x^2

it's asking where dy/dx = 3

.'. 3x^2=3

x^2 = 1

x = +-1

then sub x-value into original eq and find y-value and you've got your points

 

[Smile12345]

dy/dx = 3x^2

it's asking where dy/dx = 3

.'. 3x^2=3

x^2 = 1

x = +-1

then sub x-value into original eq and find y-value and you've got your points

Thanks heaps...

How about this.. The tangent at point P on the curve y=4x^2 + 1 is parallel to the x axis. Find the coordinates of P.

Thanks in advance. It's just (0,1) isn't it? I'm thinking 'too hard'...