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

y= -9

Extend the brackets first.

The gradient of the tangent can be found by differentiating f(x) and evaluating f'(2).

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

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

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The equation of the tangent line of f(x)=(x-5)(x+1) at x=2 is y = -3x + 7.

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