What is the tangent of f(x)=x^2+4,\at x=1 ?

The tangent of f(x)=x^2+4,\at x=1 is y=2x+3

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tangent of f(x)=x^2+4,\at x=1

tangent of

f (x) = x^{2} + 4, at x = 1

y = 2 x + 3

Solution steps

Find the tangent point:

(1, 5)

Find the slope of

f (x) = x^{2} + 4 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} + 4

x = 1 : m = 2

Find the line with slope m=

2

and passing through

(1, 5) : y = 2 x + 3

y = 2 x + 3

f (x) = x^{5} - 3 (x)^{3} + 6

6 x^{5} (x^{4} - 8)

\frac{( 2 s + 13 )}{s ^{2} + 4 s + 13}

\int_{0}^{4} 1

\frac{d}{d x} (7 x^{- 1})

What is the tangent of f(x)=x^2+4,\at x=1 ?
The tangent of f(x)=x^2+4,\at x=1 is y=2x+3