What is the tangent of f(x)=3-4x^2,(4,-61) ?

The tangent of f(x)=3-4x^2,(4,-61) is y=-32x+67

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tangent of f(x)=3-4x^2,(4,-61)

tangent of

f (x) = 3 - 4 x^{2}, (4, - 61)

y = - 32 x + 67

Solution steps

Find the slope of

f (x) = 3 - 4 x^{2} : \frac{df ( x )}{d x} = - 8 x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 32

Find the line with slope m=

- 32

and passing through

(4, - 61) : y = - 32 x + 67

y = - 32 x + 67

5 cos (t)

\frac{d}{d x} (3 e^{x} sin (x))

y^{^{''}} + 16 y = - 16 sec^{2} (4 x)

\frac{\partial}{\partial x} (\frac{c + 1}{c + x})

\frac{d}{d x} (\frac{x - 3}{x + 2})

What is the tangent of f(x)=3-4x^2,(4,-61) ?
The tangent of f(x)=3-4x^2,(4,-61) is y=-32x+67