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

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

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tangent of f(x)=x^2+x,(1,2)

tangent of

f (x) = x^{2} + x, (1, 2)

y = 3 x - 1

Solution steps

Find the slope of

f (x) = x^{2} + x : \frac{df ( x )}{d x} = 2 x + 1

Compute the slope of

f (x) = x^{2} + x

(1, 2) : m = 3

Find the line with slope m=

3

and passing through

(1, 2) : y = 3 x - 1

y = 3 x - 1

(- 1, - 4), (0, 2)

- x^{2} + 3 x + 1, - x - 1, [0, 4]

\int \frac{- 3}{x} d x

\int \frac{4 t - 3}{t + 1} d t

f (x) = x^{4} + 5 x^{2} - x, (1, 5)

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