What is the tangent of f(x)=x^2+2,(-3,11) ?

The tangent of f(x)=x^2+2,(-3,11) is y=-6x-7

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tangent of f(x)=x^2+2,(-3,11)

tangent of

f (x) = x^{2} + 2, (- 3, 11)

y = - 6 x - 7

Solution steps

Find the slope of

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

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 = - 6

Find the line with slope m=

- 6

and passing through

(- 3, 11) : y = - 6 x - 7

y = - 6 x - 7

\frac{\partial}{\partial x} (x e^{y} + y e^{z} + z e^{x})

y (1 + x^{2}) y^{^{'}} - x (1 + y^{2}) = 0

y = 12

x \to 0 lim (2 x - 7)

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

What is the tangent of f(x)=x^2+2,(-3,11) ?
The tangent of f(x)=x^2+2,(-3,11) is y=-6x-7