Asked by John
Find the equation of the tangent line to the curve y=e^2x + 4x at the point (0,1)
Answers
Answered by
Reiny
dy/dx = 2 e^(2x) + 4
at the point (0,1)
dy/dx = 2e^0 + 4
= 6
y = 6x + 1
since (0,1) is the y-intercept we can just state the answer.
btw, one would expect a student to correctly spell the subject he/she is studying.
at the point (0,1)
dy/dx = 2e^0 + 4
= 6
y = 6x + 1
since (0,1) is the y-intercept we can just state the answer.
btw, one would expect a student to correctly spell the subject he/she is studying.
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