Find the point where the tangent is horizontal for y = (5x+2)^2(1-3x)^3

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justaguide | College Teacher | (Level 2) Distinguished Educator

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For a function f(x), the slope of the tangent at any point x = c is the value of f'(c).

The function y = (5x+2)^2(1-3x)^3

y' = 10*(5x + 2)(1 - 3x)^3 - 9*(5x+2)^2*(1-3x)^2

A horizontal asymptote has a slope equal to 0.

10*(5x + 2)(1 - 3x)^3 - 9*(5x+2)^2*(1-3x)^2 = 0

=> (5x + 2)(1 - 3x)^2*(10*(1 - 3x) - 9(5x + 2)) = 0

=> (5x + 2)(1 - 3x)^2*(10 - 30x - 45x - 18) = 0

=> (5x + 2)(1 - 3x)^2*(-75x - 8) = 0

=> x = -2/5, x = 1/3 and x = -8/75

The tangents are horizontal at the points where x = -2/5, x = 1/3 and x = -8/75

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to remember that the tangent line to a graph is horizontal if the angle beween the tangent line and x axis is 0 or the slope of tangent line is parallel to x axis.

If the tangent line to the graph of a function is parallel to x axis, then the tangency point expresses a critical point of the function, either a maximum or a minimum point.

You need to differentiate the function and then you need to solve the equation f'(x)=0 such that:

f'(x) = 2*5(5x+2)(1-3x)^3 - 9(5x+2)^2(1-3x)^2

You need to factor out (5x+2)(1-3x)^2 such that:

f'(x) = (5x+2)(1-3x)^2*(10 - 30x - 45x - 18)

Collecting like terms yields:

f'(x) = (5x+2)(1-3x)^2*(-8 - 75x)

You need to solve the equation f'(x) = 0 to find the critical values of function such that:

(5x+2)(1-3x)^2*(-8 - 75x) = 0

5x + 2 = 0 => x = -2/5

1 - 3x = 0 => x = 1/3

-8-75x = 0 => x = -8/75

Hence, the tangent lines is parallel to x axis at the critical values x = -8/75; x = -2/5 ; x = 1/3 as the graph below shows:

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