What is the tangent of f(x)=x^2,(-1,-3) ?

The tangent of f(x)=x^2,(-1,-3) is y=2x-1 \lor y=-6x-9

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tangent of f(x)=x^2,(-1,-3)

tangent of

f (x) = x^{2}, (- 1, - 3)

y = 2 x - 1 or y = - 6 x - 9

Solution steps

Slope of the tangent line at

(- 1, - 3) : m = \frac{x ^{2} + 3}{x + 1}

Find the slope of

f (x) = x^{2} : m = 2 x

Solve

2 x = \frac{x ^{2} + 3}{x + 1} : x = 1, x = - 3

Find the tangent line at

x = 1 : y = 2 x - 1

Find the tangent line at

x = - 3 : y = - 6 x - 9

y = 2 x - 1 or y = - 6 x - 9

x^{2} - 4 x, x

x \to 3 lim (5 (x - 3) ln (x - 3))

f (x) = \frac{1}{x} + 3, at x = - 1

x \to 5 lim (- \frac{9}{x + 5})

\int_{1}^{3} \frac{1}{3 - x} d x

What is the tangent of f(x)=x^2,(-1,-3) ?
The tangent of f(x)=x^2,(-1,-3) is y=2x-1 \lor y=-6x-9