What is the tangent of f(x)=x^2+9,(-5,34) ?

The tangent of f(x)=x^2+9,(-5,34) is y=-10x-16

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tangent of f(x)=x^2+9,(-5,34)

tangent of

f (x) = x^{2} + 9, (- 5, 34)

y = - 10 x - 16

Solution steps

Find the slope of

f (x) = x^{2} + 9 : \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 = - 10

Find the line with slope m=

- 10

and passing through

(- 5, 34) : y = - 10 x - 16

y = - 10 x - 16

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

x \to 0 lim (\frac{3}{x} - \frac{2}{ln ( x )})

\int 2 π (4 x^{2} + 4 x + 1) d x

x \to \frac{2}{3} lim (\frac{27 x ^{3} - 8}{9 x ^{2} - 4})

d er i v a t i v e f (x) = \frac{cos ( x )}{x ^{3}}

What is the tangent of f(x)=x^2+9,(-5,34) ?
The tangent of f(x)=x^2+9,(-5,34) is y=-10x-16