What is the tangent of x^2+5x,\at x=4 ?

The tangent of x^2+5x,\at x=4 is y=13x-16

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tangent of x^2+5x,\at x=4

tangent of

x^{2} + 5 x, at x = 4

y = 13 x - 16

Solution steps

Find the tangent point:

(4, 36)

Find the slope of

y = x^{2} + 5 x : \frac{d y}{d x} = 2 x + 5

Compute the slope of

y = x^{2} + 5 x

x = 4 : m = 13

Find the line with slope m=

13

and passing through

(4, 36) : y = 13 x - 16

y = 13 x - 16

x \to + 4 + lim (\frac{7}{x - 4})

\int 1 - 25 x^{2} d x

\int \frac{20 x ^{4}}{4 x ^{5} + 3} d x

\int 3 x (x^{2} + 2)^{5} d x

\frac{d}{d x} (\frac{x ^{2} - 3 x}{x + 1})

What is the tangent of x^2+5x,\at x=4 ?
The tangent of x^2+5x,\at x=4 is y=13x-16