Find the equation of the tangent to the curve y=x^3-4x^2+2 at the point (3,-7)

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y=x3-4x2+2

A tangent to a curve at a specific point along the line will have excatly the same gradient as the curve at that point. For example if the point was (0,0) the tangent would just be a horizontal line along the x-axis. To work out the gradient we simply differentiate the curve, set this to equal 0 and solve with the given value of x.

Here dy/dx = 3x2-8x

Setting this to 0 and solving gives us 3(3)2-8(3)=3 and so our gradient at the point (3,-7) is 3

We are now able to use the equation y-y(1)=m(x-x(1)) where m is the gradient and x(1) & y(1) are the coordinates we've been given.

Rearranging we get y=3(x-3)-7

y=3x-16 and this is the equation of the tangent to the curve y=x3-4x2+2 at the point (3,-7)

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