Question

What is the equation of the tangent line of `f(x)=(x-5)(x+1)` at `x=2`?

Answer 1

y= -9

Explanation:

Begin by expanding the brackets.

`rArrf(x)= (x-5)(x+1) = x^2 -4x - 5`

differentiating f(x) and evaluating f'(2) will give the value of the gradient of the tangent.

`rArr f'(x) = 2x-4 " and " f'(2) = 4-4=0`

Since f'(2)=0 , then the tangent is parallel to the x-axis with equation y = c

now f(2) =`(2)^2 -4(2) - 5 = -9`

The point (2 , -9) is therefore on the tangent and it's equation is
y = -9
`"-----------------------------------------------------------------------"`
Since f(x) is a quadratic function this could have been approached using algebraic method.

By completing the square on `x^2-4x-5" we obtain "`

`x^2-4x-5 = (x-2)^2 -4 -5 = (x-2)^2 -9`

which has a vertex at (2 ,-9) and hence equation of tangent is
y = -9
graph{x^2-4x-5 [-20, 20, -10, 10]}