What is the tangent of f(x)=4-3x^2,(5,-71) ?

The tangent of f(x)=4-3x^2,(5,-71) is y=-30x+79

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tangent of f(x)=4-3x^2,(5,-71)

tangent of

f (x) = 4 - 3 x^{2}, (5, - 71)

y = - 30 x + 79

Solution steps

Find the slope of

f (x) = 4 - 3 x^{2} : \frac{df ( x )}{d x} = - 6 x

Compute the slope of

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

(5, - 71) : m = - 30

Find the line with slope m=

- 30

and passing through

(5, - 71) : y = - 30 x + 79

y = - 30 x + 79

\frac{- 0.5}{s ^{2} + s + 1}

\int x - 2 + \frac{3}{( 3 x - 5 ) ^{6}} d x

\frac{t}{t ^{2} + 8}

(2 x - 7) 4 (x^{2} + x + 1) 5

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

What is the tangent of f(x)=4-3x^2,(5,-71) ?
The tangent of f(x)=4-3x^2,(5,-71) is y=-30x+79