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Find the coordinates of points where the graph of f(x) has horizontal tangents. As a check, graph...

Question:

Find the coordinates of points where the graph of f(x) has horizontal tangents. As a check, graph f(x) and see whether the points you found look as though they have horizontal tangents.

{eq}f(x) = -x^3 + 6x^2 - 9x + 13 {/eq}

Horizontal Tangent Line

The slope of a tangent line to a function is the value of the derivative at a point. Since the slope of a horizontal line is zero, a horizontal tangent line occurs when the derivative is equal to zero. These points are also called critical points.

Answer and Explanation:

In order to find where this function has horizontal tangent lines, we need to first differentiate this function. This is a polynomial, which can be differentiated by applying the power rule.

{eq}f'(x) = -3x^2 + 12x - 9 {/eq}


A horizontal line has a slope of zero, and since the slope of a tangent line comes from the derivative, we can find where the function has horizontal tangent lines by finding when the derivative equals zero.

{eq}-3x^2 + 12x - 9 = 0\\ -3(x^2 - 4x + 3) = 0\\ -3(x-3)(x-1) = 0\\ x = 3\\ x = 1 {/eq}


Let's graph this function to show that these are the two points where this function has a horizontal tangent line.

The instantaneous slope is indeed zero at both 1 and 3.

Graph of -x^3 + 6x^2 - 9x + 13


Learn more about this topic:

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Tangent Line: Definition & Equation

from NY Regents Exam - Geometry: Tutoring Solution

Chapter 1 / Lesson 11
20K

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