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

The tangent of f(x)=x^2+11x-15,\at x=1 is y=13x-16

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tangent of f(x)=x^2+11x-15,\at x=1

tangent of

f (x) = x^{2} + 11 x - 15, at x = 1

y = 13 x - 16

Solution steps

Find the tangent point:

(1, - 3)

Find the slope of

f (x) = x^{2} + 11 x - 15 : \frac{df ( x )}{d x} = 2 x + 11

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 = 13

Find the line with slope m=

13

and passing through

(1, - 3) : y = 13 x - 16

y = 13 x - 16

\int 2 x^{3} \cdot cos (x^{2}) d x

\frac{d}{d x} (- 4 x - 3)

x y^{^{'}} + 2 y = 2 x^{2} - 3 x

\int 1.11 x e^{(2.76 x)} d x

\frac{d}{d x} ((\frac{1}{x})^{x})

What is the tangent of f(x)=x^2+11x-15,\at x=1 ?
The tangent of f(x)=x^2+11x-15,\at x=1 is y=13x-16