# How do you find the equation of the tangent line to #p(t) = 4 t^3 + 2 t + 5# at t=2?

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To find the equation of the tangent line to the function p(t) = 4t^3 + 2t + 5 at t = 2, we need to find the derivative of the function and evaluate it at t = 2. The derivative of p(t) is given by p'(t) = 12t^2 + 2. Evaluating p'(t) at t = 2, we get p'(2) = 12(2)^2 + 2 = 52. Therefore, the slope of the tangent line is 52.

To find the y-intercept of the tangent line, we substitute the values of t and p(t) into the equation of the tangent line, y - p(2) = m(t - 2), where m is the slope. Substituting t = 2, p(2) = 4(2)^3 + 2(2) + 5 = 49, and m = 52, we have y - 49 = 52(t - 2). Simplifying this equation gives y = 52t - 105.

Therefore, the equation of the tangent line to p(t) = 4t^3 + 2t + 5 at t = 2 is y = 52t - 105.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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