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

The tangent of f(x)=x^2+5x-2,\at x=4 is y=13x-18

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tangent of f(x)=x^2+5x-2,\at x=4

tangent of

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

y = 13 x - 18

Solution steps

Find the tangent point:

(4, 34)

Find the slope of

f (x) = x^{2} + 5 x - 2 : \frac{df ( x )}{d x} = 2 x + 5

Compute the slope of

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

x = 4 : m = 13

Find the line with slope m=

13

and passing through

(4, 34) : y = 13 x - 18

y = 13 x - 18

x \to 0 lim (2 x^{2} + 3 x)

\int \frac{1 + 4 x}{1 + x + 2 x ^{2}} d x

\frac{d}{d x} (\frac{2 x ^{2} - 4 x + 3}{2 - 3 x})

\int f (r) d r

\int \frac{t}{t ^{4} + 6} d t

What is the tangent of f(x)=x^2+5x-2,\at x=4 ?
The tangent of f(x)=x^2+5x-2,\at x=4 is y=13x-18