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

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

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

tangent of

f (x) = 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

f (x) = 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

x \to 8 + lim (\frac{1}{( x - 8 )})

(7 x + 3 y) d x + (3 x - 8 y 3) d y = 0

\frac{k x + a}{b x + c}

\frac{\partial}{\partial z} (z sin (y))

\int_{0}^{3} \frac{5 t}{( t - 4 ) ^{2}} d t

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