What is the tangent of f(x)=x^2+5,(-3,14) ?

The tangent of f(x)=x^2+5,(-3,14) is y=-6x-4

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tangent of f(x)=x^2+5,(-3,14)

tangent of

f (x) = x^{2} + 5, (- 3, 14)

y = - 6 x - 4

Solution steps

Find the slope of

f (x) = x^{2} + 5 : \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, 14) : y = - 6 x - 4

y = - 6 x - 4

(3 x^{2})^{^{'}}

\int \frac{( 6 x - 1 )}{x ^{3} ( 2 x - 1 )} d x

4 a x - x

f (x) = \frac{5 x + 1}{5 x ^{2} + 1}

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

What is the tangent of f(x)=x^2+5,(-3,14) ?
The tangent of f(x)=x^2+5,(-3,14) is y=-6x-4