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

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

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

tangent of

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

y = 2 x + 1

Solution steps

Find the slope of

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

Compute the slope of

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

(1, 3) : m = 2

Find the line with slope m=

2

and passing through

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

y = 2 x + 1

n = 1 \sum \infty (\frac{15}{4})^{n + 1}

\frac{\partial}{\partial x} (x^{y^{3}})

9 x - \frac{7}{x}

x \to - 9 lim (\frac{1}{( x + 9 ) ^{2}})

\int cos^{3} (3 x) sin^{- 2} (3 x) d x

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