What is the tangent of f(x)=x^2+8,(1,9) ?

The tangent of f(x)=x^2+8,(1,9) is y=2x+7

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tangent of f(x)=x^2+8,(1,9)

tangent of

f (x) = x^{2} + 8, (1, 9)

y = 2 x + 7

Solution steps

Find the slope of

f (x) = x^{2} + 8 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} + 8

(1, 9) : m = 2

Find the line with slope m=

2

and passing through

(1, 9) : y = 2 x + 7

y = 2 x + 7

\frac{d}{d x} (\frac{4 x ^{3}}{x ^{4} + 27})

\frac{d}{d x} (arcsin (tan (x^{2} + 1)))

\frac{\partial}{\partial x} (ln (x^{8} y))

x \to 0 lim (- \frac{1}{x ^{2}})

\frac{d}{d x} (3 sin^{2} (2 x))

What is the tangent of f(x)=x^2+8,(1,9) ?
The tangent of f(x)=x^2+8,(1,9) is y=2x+7