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

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

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tangent of x^2+3,\at (x=1)

tangent of

x^{2} + 3, at (x = 1)

y = 2 x + 2

Solution steps

Find the tangent point:

(1, 4)

Find the slope of

y = x^{2} + 3 : \frac{d y}{d x} = 2 x

Compute the slope of

y = x^{2} + 3

x = 1 : m = 2

Find the line with slope m=

2

and passing through

(1, 4) : y = 2 x + 2

y = 2 x + 2

\frac{s ^{2} - 1}{s ^{4} + 3 s ^{2} + 2}

\int \frac{1}{2 x ^{2} + 2} d x

- t

\frac{\partial}{\partial x} (- 2 s t)

x \to 0 + lim (\frac{x}{cos ( 4 x ) - 1})

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