What is the tangent of y=x^2+4,(-2,8) ?

The tangent of y=x^2+4,(-2,8) is y=-4x

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tangent of y=x^2+4,(-2,8)

tangent of

y = x^{2} + 4, (- 2, 8)

y = - 4 x

Solution steps

Find the slope of

y = x^{2} + 4 : \frac{d y}{d x} = 2 x

Compute the slope of

y = x^{2} + 4

(- 2, 8) : m = - 4

Find the line with slope m=

- 4

and passing through

(- 2, 8) : y = - 4 x

y = - 4 x

f (x) = 3 x^{\frac{2}{3}} - 2 x

x \to 0 - lim (- \frac{1}{x})

n = 1 \sum \infty \frac{5 n}{6 n + 5}

y = arccot (7 t)

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