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

Answer 1

Isaiah Allen

y= -9

Extend the brackets first.

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

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

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

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

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#

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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Answer 2

Scarlett Allison

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

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Answer from HIX Tutor

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