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What is the slope of #f(t) = (t-2t^2,t)# at #t =-1#?

Answer 1

William Abdalla

Slope at #t=-1# is #1/5#

The slope of a function #f(x)# is given by its derivative #f'(x)#.

In a parametric equation of type #f(t)=(x(t),y(t))#, #f'(t)=((dy)/(dt))/((dx)/(dt))#

As #(dy)/(dt)=1# and #(dx)/(dt)=1-4t#

#f'(t)=1/(1-4t)#

and slope at #t=-1# is #1/5# and this at #(-3,-1)#

graph{(x-y+2y^2)(5y+5-x-3)=0 [-5.813, 4.187, -2.68, 2.32]}

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Answer 2

Aurora Alvarado

To find the slope of ( f(t) = (t - 2t^2, t) ) at ( t = -1 ), we first find the derivative of ( f(t) ) with respect to ( t ), and then evaluate it at ( t = -1 ).

[ f(t) = (t - 2t^2, t) ]

[ \frac{df}{dt} = \left( \frac{d(t - 2t^2)}{dt}, \frac{dt}{dt} \right) ]

[ \frac{df}{dt} = (1 - 4t, 1) ]

Now, evaluate the derivative at ( t = -1 ):

[ \frac{df}{dt} \bigg|_{t = -1} = (1 - 4(-1), 1) ]

[ \frac{df}{dt} \bigg|_{t = -1} = (1 + 4, 1) ]

[ \frac{df}{dt} \bigg|_{t = -1} = (5, 1) ]

Therefore, the slope of ( f(t) = (t - 2t^2, t) ) at ( t = -1 ) is ( m = 5 ).

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Answer from HIX Tutor

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