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

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

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

tangent of

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

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

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 = 3, x = - 1

Find the tangent line at

x = 3 : y = 6 x - 9

Find the tangent line at

x = - 1 : y = - 2 x - 1

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

\frac{\partial}{\partial x} (x^{2} sin (k x))

\frac{d}{d x} (e^{x} + e^{- 2 x})

\int_{0}^{π} sin (x) d x

y^{^{''}} + y^{^{'}} - 2 y = x^{3} - e^{- x}

\frac{d}{d x} (cos (x) + ln (tan (\frac{x}{2})))

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