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Find an equation of the tangent line to the graph of the function at the given point.y = (e4x − 4)2,    (0, 9)y =

Question

Find an equation of the tangent line to the graph of the function at the given point.

y = (e4x − 4)2,    (0, 9)
y =
check_circleAnswer
Step 1

We first have to find y'

We apply exponent rule and chain rule.

y'= 8(e^(4x))*(e^(4x)) -4)

y (c44)
2
y'-2(e -4)2-1
-(e 4-4)
dx
(et(4x)-0)
y'-2(e -4)
y'2(e -4) (e (4)
y' 8e*(e 4-4)
help_outline

Image Transcriptionclose

y (c44) 2 y'-2(e -4)2-1 -(e 4-4) dx (et(4x)-0) y'-2(e -4) y'2(e -4) (e (4) y' 8e*(e 4-4)

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Step 2

At x=0 , we calculate y'

That will be the slope ...

У-38e " (e*-4)
4(0)
У'—8e
у-8e (е°-4)
у-38(1)(1-4)
у38(-3)
у3-24
help_outline

Image Transcriptionclose

У-38e " (e*-4) 4(0) У'—8e у-8e (е°-4) у-38(1)(1-4) у38(-3) у3-24

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Tagged in

Math

Calculus

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