What is the tangent of y=x^2+x,(2,-3) ?

The tangent of y=x^2+x,(2,-3) is y=11x-25 \lor y=-x-1

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tangent of y=x^2+x,(2,-3)

tangent of

y = x^{2} + x, (2, - 3)

y = 11 x - 25 or y = - x - 1

Solution steps

Slope of the tangent line at

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

Find the slope of

y = x^{2} + x : m = 2 x + 1

Solve

2 x + 1 = \frac{x ^{2} + x + 3}{x - 2} : x = 5, x = - 1

Find the tangent line at

x = 5 : y = 11 x - 25

Find the tangent line at

x = - 1 : y = - x - 1

y = 11 x - 25 or y = - x - 1

f^{^{'}} (x) = x^{3} (x^{4} - 1)

x \to 0 lim (\frac{1}{1 + 5 ^{\frac{2}{x}}})

\int sin (\frac{1}{3} x) d x

\frac{\partial}{\partial x} (2 sin (3 x))

\frac{d}{d x} (cos^{3} (3 x))

What is the tangent of y=x^2+x,(2,-3) ?
The tangent of y=x^2+x,(2,-3) is y=11x-25 \lor y=-x-1