By A. C. Burdette (Auth.)

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Radius - 3 —è (-3,-1) Fig. 3-9 and the radius is 3. This, together with the coordinates of the center, sub stituted in (3-2) gives us the required equation (x + 3)2 + (y - 2) 2 = 9, or x 2 + y2 + 6x - 4y + 4 = 0. 34 3. NONLINEAR EQUATIONS AND GRAPHS Exercises 3-2 1. x + Ay - 4 = 0 x 2 + y2 + 5x = 0 2. Determine equations for the circles having the following properties: (a) Center at (2, — 5), radius 7 (b) Center at ( — 3, — 1), radius 5 (c) A diameter has the end points (5, - 4 ) and ( - 3 , 2) (d) Center at (4, —2) and tangent to x = 5 (e) Center at ( — 7, 2) and tangent to the x-axis (f) Radius 5 and tangent to the j^-axis at (0, — 1) (g) Center at (1, 3) and passing through the origin (h) Center at (0, 3) and tangent to 2x + y - 13 - 0 3.

A) Give the domain and range of φ. (b) F i n d 0 ( - 1 ) , φ(-±1φ(0), φ(\). (c) Draw the graph of φ. 11. Let f(x) = 1 — 3x. Find a function F such that f[F(x)] = x. Does F[f{x)\ = xl 4-3. Tangent to a Curve The student has already encountered the idea of a tangent to a curve in the study of circles in elementary geometry. We propose to define tangents to curves in general in a manner consistent with our experience with tangents to circles. We shall use as a guide in arriving at this definition the idea that a line L tangent to a curve C at the point P should have the same "direction" as C at P.

THE PARABOLA 37 Example 3-15. Discuss and graph x2 + 6x - 4y + 13 = 0. We first note that this equation is of the form (3-8) and therefore represents a parabola. Next we reduce it to form (3-7) by writing Ay - 13 = x2 + 6x, and then, completing the square of the right member, obtain 4y - 4 = x2 + 6x + 9, or finally 4(y - 1) = (x + 3) 2 . From this form of the equation, making use of Theorem 3-5, we conclude that the parabola extends upward from its vertex ( — 3, 1), symmetric to its axis x = — 3.