What is the tangent line to the curve y = x^3+4x+5 at the point where x = 2?

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First, we must find the value of y when x = 2.

y = x3+4x+5 = (2)3+4(2)+5 = 21

Then we must find the gradient of the tangent line. This can be done by differentiating y with respect to x and substituting x = 2.

dy/dx = 3x2+4 = 3(2)2+4 = 16

Now that we have a point (2,21) and the gradient (m = 16) of our tangent line, we can find the equation of the tangent using the formula:

y-y= m(x-x1)

y-21 = 16(x-2)

y = 16x-32+21

Thus y = 16x-11 is the equation of the tangent

Oliver T. GCSE Maths tutor, 13 plus  Maths tutor, A Level Maths tutor...

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