Question

What is the equation of the normal line of `f(x)=(x-4)e^(x-2)` at `x=0`?

Answer 1

`y = 2.463x +0.5413`

Explanation:

Equation of the line will equal
`y-y_1 = m_2(x-x_1)`

Differentiate to find the gradient function using the product rule
`dy/dx = u(dv)/dx + v(du)/dx`

`u = x-4`
`(du)/dx = 1`

`v = e^(x-2)`
`(dv)/dx = e^(x-2)`

`dy/dx = (x-4)e^(x-2)+e^(x-2)`
`dy/dx = (x-4)e^(x-2)+e^(x-2)`
`dy/dx = xe^(x-2)-3e^(x-2)`

At `x=0`
`dy/dx = -3e^(-2) = m_1`

The normal (`m_2`) is perpendicular to the tangent (`m_1`) so:
`m_2 = (-1)/(m_1)`

`m_2 = (-1)/(-3e^(-2))`
`m_2 = (1)/(3e^(-2))`
`m_2 = 2.463`

At `x=0`,
`y = (0-4)e^(0-2)`
`y = (4)e^(-2)`
`y=0.5413`

So the equation of the normal to the curve at x=0 is,
`y-y_1 = m_2(x-x_1)`
`y-0.5413 = 2.463(x-0)`
`y = 2.463x +0.5413`