What is the tangent of f(x)=x^2+5,(1,6) ?

The tangent of f(x)=x^2+5,(1,6) is y=2x+4

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tangent of f(x)=x^2+5,(1,6)

tangent of

f (x) = x^{2} + 5, (1, 6)

y = 2 x + 4

Solution steps

Find the slope of

f (x) = x^{2} + 5 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} + 5

(1, 6) : m = 2

Find the line with slope m=

2

and passing through

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

y = 2 x + 4

\frac{\partial}{\partial x} (y + x)

(2 x + 5)^{7}

\frac{\partial}{\partial x} (4 x^{2} - 2 y^{2})

\int sec (θ) d θ

(2 x + 1 + 6 y) d x + 3 d y = 0

What is the tangent of f(x)=x^2+5,(1,6) ?
The tangent of f(x)=x^2+5,(1,6) is y=2x+4