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

\frac{x}{4} - \frac{4}{x}

x^{2} \frac{d y}{d x} - 2 x y = 6 y^{4}, y (1) = \frac{1}{3}

x \to \infty lim (7 x)

\int \frac{1}{x + 7} d x

\int_{4}^{5} \frac{x}{8} 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