What is the tangent of y=-1-8x^2,(-2,-33) ?

The tangent of y=-1-8x^2,(-2,-33) is y=32x+31

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tangent of y=-1-8x^2,(-2,-33)

tangent of

y = - 1 - 8 x^{2}, (- 2, - 33)

y = 32 x + 31

Solution steps

Find the slope of

y = - 1 - 8 x^{2} : \frac{d y}{d x} = - 16 x

Compute the slope of

y = - 1 - 8 x^{2}

(- 2, - 33) : m = 32

Find the line with slope m=

32

and passing through

(- 2, - 33) : y = 32 x + 31

y = 32 x + 31

f (x) = x^{\frac{2}{5}}

n = 2 \sum \infty \frac{2}{n ( n + 2 )}

y^{^{'}} - 5 y = 2 e^{5 x}

x^{3} - 4 x

\frac{d}{d x} (7 - x)

What is the tangent of y=-1-8x^2,(-2,-33) ?
The tangent of y=-1-8x^2,(-2,-33) is y=32x+31