Question
Asked Sep 22, 2019
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m of the tangent to the curve y
2 4x2 - 2x3 at the point where x a
(a) Find the slope
=
m =
(b) Find equations of the tangent lines at the points (1, 4) and (2, 2)
(at the point (1, 4))
y(x)
(at the point (2, 2))
y(x)=
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m of the tangent to the curve y 2 4x2 - 2x3 at the point where x a (a) Find the slope = m = (b) Find equations of the tangent lines at the points (1, 4) and (2, 2) (at the point (1, 4)) y(x) (at the point (2, 2)) y(x)=

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Expert Answer

Step 1

Given,

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y 24x2 2x3 Part A Since, slope of the tangent line is derivative of y y'd) dx d(2+4x2-2x dx = 0 4 x 2x - 2 x 3x2 - 8х — бх? So, slope of the tangent to the curve at the point x a is dy) m= dx xa - 8а — ба? =

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Step 2

Part b

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Slope at point (1,4) = 8(1) - 6(1) = 8 - 6 = 2 So, equation of the tangent line at the point (1,4) is y 4 2(x 1) у — 4 %3D 2х — 2 y 2x 2 Slope at point (2,2) 8(2) 6(2) = 16 - 24 -8 So, equation of the tangent line at the point (2,2) is у — 2 %3D (-8)(х — 2) у — 2 3 —8х + 16 у %3D — 8х + 18

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Step 3

Graph of the curve and tangent at ...

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8 Curve 6 (1, 4) Tangent at (1,4) 4 (2, 2) Tangent at (2,2) 2 10 2

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Tagged in

Math

Calculus

Derivative