What is the tangent of f(x)=x^2+3x+1,\at x=4 ?

The tangent of f(x)=x^2+3x+1,\at x=4 is y=11x-15

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tangent of f(x)=x^2+3x+1,\at x=4

tangent of

f (x) = x^{2} + 3 x + 1, at x = 4

y = 11 x - 15

Solution steps

Find the tangent point:

(4, 29)

Find the slope of

f (x) = x^{2} + 3 x + 1 : \frac{df ( x )}{d x} = 2 x + 3

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 11

Find the line with slope m=

11

and passing through

(4, 29) : y = 11 x - 15

y = 11 x - 15

x \to 0 lim (ln (0))

\int x^{3} (3 + x^{4})^{5} d x

\frac{\partial}{\partial x} (2 x^{2} y + 3)

\frac{x + 1}{x - 1}

\frac{d r}{d θ} r = (4 + sec (θ)) sin (θ)

What is the tangent of f(x)=x^2+3x+1,\at x=4 ?
The tangent of f(x)=x^2+3x+1,\at x=4 is y=11x-15