What is the tangent of y=x^2-x,(4,12) ?

The tangent of y=x^2-x,(4,12) is y=7x-16

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tangent of y=x^2-x,(4,12)

tangent of

y = x^{2} - x, (4, 12)

y = 7 x - 16

Solution steps

Find the slope of

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

Compute the slope of

y = x^{2} - x

(4, 12) : m = 7

Find the line with slope m=

7

and passing through

(4, 12) : y = 7 x - 16

y = 7 x - 16

\int \frac{1}{x ^{2} ( x + 1 ) ^{3}} d x

f (x) = \frac{5 x}{x - 3}

n = 1 \sum \infty (2)^{n - 1}

\frac{1}{x ^{2} + 1} (\frac{d y}{d x}) = x e^{y}

\int \frac{4}{x ^{2} - 1} d x

What is the tangent of y=x^2-x,(4,12) ?
The tangent of y=x^2-x,(4,12) is y=7x-16