What is the tangent of f(x)=x^2-x,(-4,20) ?

The tangent of f(x)=x^2-x,(-4,20) is y=-9x-16

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tangent of f(x)=x^2-x,(-4,20)

tangent of

f (x) = x^{2} - x, (- 4, 20)

y = - 9 x - 16

Solution steps

Find the slope of

f (x) = x^{2} - x : \frac{df ( x )}{d x} = 2 x - 1

Compute the slope of

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

(- 4, 20) : m = - 9

Find the line with slope m=

- 9

and passing through

(- 4, 20) : y = - 9 x - 16

y = - 9 x - 16

x \cdot y \cdot \frac{d y}{d x} - y^{2} = 2 \cdot x^{2}

n = 4 \sum \infty \frac{1}{n ( n + 1 )}

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

\int_{0}^{2} \frac{1}{x + 3} d x

cos (2 y)

What is the tangent of f(x)=x^2-x,(-4,20) ?
The tangent of f(x)=x^2-x,(-4,20) is y=-9x-16