What is the tangent of f(x)=5-8x^2,(-2,-27) ?

The tangent of f(x)=5-8x^2,(-2,-27) is y=32x+37

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tangent of f(x)=5-8x^2,(-2,-27)

tangent of

f (x) = 5 - 8 x^{2}, (- 2, - 27)

y = 32 x + 37

Solution steps

Find the slope of

f (x) = 5 - 8 x^{2} : \frac{df ( x )}{d x} = - 16 x

Compute the slope of

f (x) = 5 - 8 x^{2}

(- 2, - 27) : m = 32

Find the line with slope m=

32

and passing through

(- 2, - 27) : y = 32 x + 37

y = 32 x + 37

(x + 2 y) d x + (2 x - y) d y = 0

y^{^{''}} - 3 y^{^{'}} + 2 y = 0

x + y y^{^{'}} = 0

\frac{d y}{d x} = \frac{( e ^{x} )}{y ^{2}}

\int \frac{1}{p ^{2} + 1} d p

What is the tangent of f(x)=5-8x^2,(-2,-27) ?
The tangent of f(x)=5-8x^2,(-2,-27) is y=32x+37