AP Calculus ABeasymcq1 pt

Consider the curve given by the equation $x^2 + y^2 = 2x + 4y - 4$ . At what points on the curve is the tangent line horizontal?

(1, 1)

and

(1, 3)

(-1, 1)

and

(-1, 3)

(1, -1)

and

(1, -3)

(-1, -1)

and

(-1, -3)

Explanation

Core Concept

Correct. By completing the square, the equation can be rewritten as $(x-1)^2 + (y-2)^2 = 1$ , which is a circle centered at $(1, 2)$ with a radius of 1. The horizontal tangent lines of a circle occur at the points directly above and below the center, which are $(1, 1)$ and $(1, 3)$ . Equivalently, using implicit differentiation yields $y' = (1-x)/(y-2)$ , which is zero when $x = 1$ ; substituting $x = 1$ into the original equation yields $y = 1$ and $y = 3$ .

Correct Answer

A $(1, 1)$ and $(1, 3)$

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Practice Unit 5: Analytical Applications of Differentiation Questions →

Consider the curve given by the equation x2+y2=2x+4y−4x^2 + y^2 = 2x + 4y - 4x2+y2=2x+4y−4. At what points on the curve is the tangent line horizontal?

Explanation

Core Concept

Correct Answer

More Unit 5: Analytical Applications of Differentiation practice questions

Consider the curve given by the equation $x^2 + y^2 = 2x + 4y - 4$ . At what points on the curve is the tangent line horizontal?