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How do you find the equation of the line tangent to the graph of #f(t) = t − 13 t^2#, a = 3?

Answer 1

Christopher Agresta

It is #y=-77x+87#

The general equation is

#y-f(a)=f'(a)(x-a)#

Hence we have that

#f(a)=f(3)=-114#

#f'(a)=1-26a=1-26*3=-77#

#y-f(a)=f'(a)(x-a)=>y-(-144)=(-77)(x-3)=>y+144=-77x+231=> y=-77x+87#

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Answer 2

Ezra Adams

To find the equation of the line tangent to the graph of f(t) = t - 13t^2 at a = 3, we need to find the derivative of f(t) and evaluate it at t = 3.

The derivative of f(t) is f'(t) = 1 - 26t.

To find the slope of the tangent line, we substitute t = 3 into f'(t): f'(3) = 1 - 26(3) = -77.

The slope of the tangent line is -77.

Using the point-slope form of a linear equation, the equation of the tangent line is y - f(3) = -77(t - 3).

Simplifying, we get y - (-90) = -77t + 231.

The equation of the tangent line is y = -77t + 321.

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Answer from HIX Tutor

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.