What is the tangent of f(x)=x^2+7,16,\at x=3 ?

The tangent of f(x)=x^2+7,16,\at x=3 is y=6x-2

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tangent of f(x)=x^2+7,16,\at x=3

tangent of

f (x) = x^{2} + 7, 16, at x = 3

y = 6 x - 2

Solution steps

Find the tangent point:

(3, 16)

Find the slope of

f (x) = x^{2} + 7 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} + 7

x = 3 : m = 6

Find the line with slope m=

6

and passing through

(3, 16) : y = 6 x - 2

y = 6 x - 2

\int \frac{e ^{- 3 t}}{t} d t

\frac{1}{3} (3 - 5 x)^{3}

y y^{^{'}} + 4 x = 0

x \to 2 lim (\frac{x ^{3} + 8}{x + 2})

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

What is the tangent of f(x)=x^2+7,16,\at x=3 ?
The tangent of f(x)=x^2+7,16,\at x=3 is y=6x-2