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

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

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

tangent of

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

y = 2 x + 7

Solution steps

Find the tangent point:

(1, 9)

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

x = 1 : m = 2

Find the line with slope m=

2

and passing through

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

y = 2 x + 7

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

(e^{x} + 1)^{2}

x \to 0 lim (\frac{x ^{2} + 3}{x ^{4}})

\int_{0}^{1} 5 cos (x^{2}) d x

x \to 0 lim (\frac{( 6 ^{x} - 8 ^{x} )}{x})

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