What is the slope of the tangent to the curve y = 18x^2 + 3x - 1 at the points where it intersects the x-axis

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The slope of the tangent to the curve y = 18x^2 + 3x - 1 at the points where it intersects the x-axis has to be determined.

At the points where the curve intersects the x-axis, the value of y = 0.

18x^2 + 3x - 1 = 0

=> 18x^2 + 6x - 3x - 1 = 0

=> 6x(3x + 1) - 1(3x + 1) = 0

=> (6x - 1)(3x + 1) = 0

=> x = 1/6 and x = -1/3

The slope of the tangent to the the curve at any point is equal to the value of y' at that point

y = 18x^2 + 3x - 1

=> y' = 36x + 3

At x = 1/6, y' = 9 and at x = -1/3, y' = -9

The slope of the tangent to the curve y = 18x^2 + 3x - 1 at the points where it intersects the x-axis is 9 and -9 respectively.

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