What is the tangent of y=(x-4)^2,\at x=5 ?

The tangent of y=(x-4)^2,\at x=5 is y=2x-9

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tangent of y=(x-4)^2,\at x=5

tangent of

y = (x - 4)^{2}, at x = 5

y = 2 x - 9

Solution steps

Find the tangent point:

(5, 1)

Find the slope of

y = (x - 4)^{2} : \frac{d y}{d x} = 2 (x - 4)

Compute the slope of

y = (x - 4)^{2}

x = 5 : m = 2

Find the line with slope m=

2

and passing through

(5, 1) : y = 2 x - 9

y = 2 x - 9

\frac{d y}{d t} = - 2 y + 10

\int x^{3} (x^{4} - 1)^{5} d x

y = 4 e^{x} cos (x), (0, 4)

\frac{\partial}{\partial x} (x \cdot e^{y} + ln (x) \cdot y)

\frac{v}{v + \frac{c}{v}}

What is the tangent of y=(x-4)^2,\at x=5 ?
The tangent of y=(x-4)^2,\at x=5 is y=2x-9