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Calculus help please

Calculus help please-example-1
User Caity
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Explanation:

THE POINT WHERE


f( - 4) = ( { - 4})^(2) + 1 \\ f( - 4) = 16 + 1 \\ f( - 4) = 17

Now we have point (-4,17)

the gradient of the tangent line at that point of f(x) is the first derivative f'(x)=2x at that point

f'(x) =m

f'(-4)=2(-4)

f'(-4)=-8

The gradient therefore is -8

the general equation of a straight line is given by


y = mx + c \\ 17 = ( - 8)( - 4) + c \\ 17 = 32 + c \\ c = 17 - 32 \\ c = - 15 \\ \\ y = - 8x - 15

HOPE THIS HELPS!!

User Zvi Mints
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