By H. Keisler

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The graphs of all power functions y = x" (starting with n = 2) touch the x-axis at the origin of coordinates, approaching it more rapidly the larger n is (Fig. 16). I. Construct the graph of the function y (h) factoring the polynomial =: (I + a)(I + a) ... \ ... 027: ... , b,,: 100: 400: 900; .... 3. 1 = lOx: (b) x 3 (d) x 5 = x + I; - x - I = O. 4. \···\·a+···. ~ x2 Can the first sequence catch up with the second (that is, can the inequality a" > b" be satisfied for some II)? • For negative values of x the situation is analogous.

Translate the parabola y = x 2 along the x-axis so that it will pass through the point (3, 2). The graph of what function is obtained (Fig. 12)? + + H Let us now see what can be said about the solution of the quadratic equation x 2 px + q = 0 using the graph of the function y = x 2 + px + q. , " x + 49 Fig. 12 _------------------- x -um--" The roots of this equation are those values of x for which the value of the function y = x 2 + px + q is equal to zero. rn the graph these points have ordinates equal to zero; that is, they lie on the x-axis.

An analogous picture results near the straight line x = -I: (e) y = 0 at x = 0; the curve passes through the origin (Fig. 18). (d) For numbers large in absolute value, both terms are small in absolute value, and both extreme branches of the graph approach the x-axis: the right from above, and the left from below (Fig. 19). Combining all this information, we can obtain the general form of the graph (Fig. 20). Show that this graph is symmetric with respect to the origin. o Fig. \ 1 I I I 1 I I I I I • I , I I I o I 1 87 X I I I I I II \1 ~ig.