An equation of the tangent line to the curve y= 4x2-8x-5 at the point (-1, 5) isO y14x-9Oy 14x71O y-14x10O y 8x-6O y (8x-6x- 1) + 5
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An equation of the tangent line to the curve y= 4x2-8x-5 at the point (-1, 5) is O y14x-9 Oy 14x71 O y-14x10 O y 8x-6 O y (8x-6x- 1) + 5
The curve is given by,
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y 4x26x-5 at the point 1,5
Differentiate y with respect to x,
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dy d (4x-6x -5) dx dx d (4x2)-^(6x)-(5) dx dx dx =8x -6
Find the value of the derivative ...
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dy = { 8(-1)6 dx -8- 6 -14
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Calculus
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