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Code Scholars Tutoring · AP Calculus BC

Reading derivative graphs and implicit curves

Unit 5 · Topics 5.8, 5.9, 5.12 · Detailed notes, a worked example, and a self-check with reasoning.

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Understand the idea

Translate graph evidence carefully: the sign of the derivative controls the direction of the function; the direction of the derivative controls the concavity of the function. Zeros on a derivative graph are candidates for horizontal tangents. For an implicit curve find horizontal tangents by setting the numerator of the slope to zero with a nonzero denominator, and reverse the roles for vertical-tangent candidates. Verify points on the original curve.

y′=−xyon x2+y2=r2 when y≠0y'=-\frac xy\quad\text{on }x^2+y^2=r^2\text{ when }y\ne0

Worked example

If a graph of f′\displaystyle f' lies below the axis and rises on (2,5)\displaystyle (2,5), describe f\displaystyle f.

f′<0\displaystyle f'<0 means f\displaystyle f is decreasing. Because f′\displaystyle f' is increasing, f\displaystyle f is concave up on that interval. The function can decrease while bending upward.

Try it yourself

Find horizontal tangents to x2+y2=9\displaystyle x^2+y^2=9.

Show solution and reasoning

The slope is −x/y\displaystyle -x/y. Set x=0\displaystyle x=0 and require y≠0\displaystyle y\ne0. The original equation gives y=±3\displaystyle y=\pm3, so the horizontal tangents occur at (0,3)\displaystyle (0,3) and (0,−3)\displaystyle (0,-3).

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