Find an equation of the horizontal tangent line to the curve given by parametric

Question

Find an equation of the horizontal tangent line to the curve given by parametric equations
x = (t^2 - t + 1), y = (2t^4 - 8t)
Please show step by step work for the problem. Please make sure your answer is correct, easy/clear to read, and has answered all parts of the question. Thanks in advance.
Find an equation of the horizontal tangent line to the curve given by parametric equations
x = (t^2 - t + 1), y = (2t^4 - 8t)
Please show step by step work for the problem. Please make sure your answer is correct, easy/clear to read, and has answered all parts of the question. Thanks in advance.

x = (t^2 - t + 1), y = (2t^4 - 8t)
Please show step by step work for the problem. Please make sure your answer is correct, easy/clear to read, and has answered all parts of the question. Thanks in advance.

Explanation / Answer

x = (t^2 - t + 1) ==>dx/dt =2t -1

, y = (2t^4 - 8t)==> dy/dt =8t3 -8

horizontal tangent occurs when dy/dt =0 given that dx/dt not equal to zero

dy/dt =0 ==>8t3 -8 =0==>8t3 =8==>t3 =1 ==>t =1

when t= 1, dx/dt =2(1)-1 =1 not equal to zero

so horizontal tangent occurs when t=1

(x,y)=((1^2 - 1 + 1),(2*1^4 - 8*1))

(x,y)=(1,- 6)

y coordinate gives horizontal tangent

equation of horizontal tangent is y =-6

or

y+6 =0

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