What is the tangent of y=x^2+4x-11,(2,-1) ?

The tangent of y=x^2+4x-11,(2,-1) is y=(8-2sqrt(2))x+4sqrt(2)-17 \lor y=(2sqrt(2)+8)x-17-4sqrt(2)

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tangent of y=x^2+4x-11,(2,-1)

tangent of

y = x^{2} + 4 x - 11, (2, - 1)

y = (8 - 22) x + 42 - 17 or y = (22 + 8) x - 17 - 42

Solution steps

Slope of the tangent line at

(2, - 1) : m = \frac{x ^{2} + 4 x - 10}{x - 2}

Find the slope of

y = x^{2} + 4 x - 11 : m = 2 x + 4

Solve

2 x + 4 = \frac{x ^{2} + 4 x - 10}{x - 2} : x = 2 - 2, x = 2 + 2

Find the tangent line at

x = 2 - 2 : y = (8 - 22) x + 42 - 17

Find the tangent line at

x = 2 + 2 : y = (22 + 8) x - 17 - 42

y = (8 - 22) x + 42 - 17 or y = (22 + 8) x - 17 - 42

f (x) = 1 + x, x = 0

\frac{d}{d x} (2 1 - x^{2})

x \to \infty lim (x^{2} + x)

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

(11 x y^{2} - 13 y) + (11 x^{2} y - 13 x) \cdot \frac{d y}{d x} = 0

What is the tangent of y=x^2+4x-11,(2,-1) ?
The tangent of y=x^2+4x-11,(2,-1) is y=(8-2sqrt(2))x+4sqrt(2)-17 \lor y=(2sqrt(2)+8)x-17-4sqrt(2)