What is the tangent of y=x^2+6x-5,\at x=2 ?

The tangent of y=x^2+6x-5,\at x=2 is y=10x-9

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tangent of y=x^2+6x-5,\at x=2

tangent of

y = x^{2} + 6 x - 5, at x = 2

y = 10 x - 9

Solution steps

Find the tangent point:

(2, 11)

Find the slope of

y = x^{2} + 6 x - 5 : \frac{d y}{d x} = 2 x + 6

Compute the slope of

y = x^{2} + 6 x - 5

x = 2 : m = 10

Find the line with slope m=

10

and passing through

(2, 11) : y = 10 x - 9

y = 10 x - 9

\int x^{9} e^{x^{10}} d x

\frac{2 d y}{d x} + 7 y = 1

\frac{1}{s - 3}

\frac{d}{d x} (\frac{ln ( x )}{2} - \frac{x ^{2}}{4})

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

What is the tangent of y=x^2+6x-5,\at x=2 ?
The tangent of y=x^2+6x-5,\at x=2 is y=10x-9