Letting Ax approach zero, the limit of (5) is m = 2ax, (6) the required rate of change, or the slope of the tangent line at any point. If x = 0, we have m = O, so that the parabola which is the graph of (3) is tangent to the x-axis at the origin. For a given value of x, the larger the value of a, the larger also is the slope of the tangent line. This justifies the statement made in Section 30 that "the larger the value of a the more rapidly the curve rises." If a = 1, so that y = x2, we have m = 2x. Hence the slope of the tangent line is less or greater than 1 according as x is less or greater than . Therefore, to the right of x = , the curve rises more rapidly than it broadens out. (Compare with the statement in Section 29). For any given value of a, the larger the value of x the larger is the value of m. Hence the curve rises more and more rapidly as it runs to the right. EXAMPLE 2. Find the slope of the line tangent at any point to the graph of y = x2 - 4x. (7) Find the slope at the point for which x = 3, construct the line, and find its equation. Find also the coördinates of the maximum or minimum points. The table of values and the graph are readily constructed. บ Passing to the limit as Ax approaches zero, the slope of the tangent line at any point is m = 2x 4. (10) Using the point-slope form of the equation of a straight line (page 66), the equation of the line tangent at A is As a check, notice that the intercept on the y-axis, - 9, agrees with the line as constructed. At a maximum or minimum point the tangent line is horizontal, and hence the slope m given by (10) must be zero. Hence Substituting this value in (7), the minimum value of y is y the minimum point is the point D(2, 4). EXERCISES 1. A ball is rolled down an inclined plane. Its distance from the starting point is given by s = t24t. Find the velocity at any time; at the instant t 3. Plot the graphs of s and v on the same axes, and from them describe the motion. = 2. Prove that a line tangent to a circle, as defined in Section 33, is perpendicular to the radius drawn to the point of contact, using the form of reasoning of elementary geometry. 3. The distance from a fixed station to a moving body is given by = t2 - 3t. Find the velocity at any time, and show that the acceleration is constant. Plot the graphs of s and v on the same axes, and describe the motion from t = 0 to t = 5. 4. Find the slope of the line tangent at any point to the graph of each of the functions below. Find the slope of the line tangent at the point for which x = 2, construct the line, and find its equation. Find the coördinates of the maximum or minimum points. (a) 4 - x2. (b) y = 3x - x2. = (c) y = x2 - 1. (d) x2 - 4y + 2x = 0. 5 Tabulate the values of the slope of the line tangent to the graph of 22 at the points for which - 3, 2, 1, 0, 1, 2, 3. What can be said of the value of m as x increases? Does the value of m, the slope of the tangent line, always increase as x increases if the graph is concave upward? Is the converse true? If a curve is concave downward, how does the slope of the tangent line change as x increases? Is the converse true? 6. Find the maximum and minimum points of y = x3- x, and then plot the graph. 7. Find the equation of the line tangent to the graph of x2 at the point for which x = 1; for which x 2; for which x 4. Find the intercept = = of each line on the y-axis. How does the intercept compare with the ordinate of the point of contact? 8. Prove that if two lines are perpendicular the slope of one is the negative reciprocal of the slope of the other. DEFINITION. The line perpendicular to a line tangent to a given curve at the point of tangency is called a normal to the curve. The slope of the normal at any point may be found from that of the tangent by the preceding exercise. 9. Find the equations of the tangent and normal to the graph of y = x2 + 2x at the point (1, 3). Construct the figure. == = 10. Find the equation of the line normal to the graph of x2 at the point for which x 1; for which x 3; for which x = 5. Find the intercept of each line on the y-axis. How does the intercept compare with the ordinate of the point at which the normal cuts the curve? 11. Find the acceleration of a moving body if its velocity is given by v = 2t2 – 5t. Plot the graphs of v and a on the same axes and discuss the variation of both functions. t3 - 9t, 12. If the position of a moving body is given by the equation s = find the velocity and acceleration. Plot the graphs of s, v, and a on the same axes. 13. Find the acceleration of a body if its position is given by s = 412 - t3. 14. If a ball is dropped the velocity after the ball has fallen s feet is given by v2 = 2gs. Find the rate of change of s with respect to v. 34. Graph of the Quadratic Function ax2 + bx + c. Denote the function by y so that In order to determine the form of the graph we shall first show that it always has a single maximum or minimum point, and then translate the axes so as to have this point for the new origin. To find the maximum or minimum point we need first the slope of the tangent line at any point. In (1) replace x by x + Ax and y by y + Ay. This gives or y + Ay = a(x + Ax)2 + b(x + Ax) + c, y + Ay = ax2 + 2ax Ax + Ax2 + bx + b Ax + c. Subtracting (1) from (2), Ay =2ax Ax + Ax2 + b Ax, (2) Passing to the limit as Ax approaches zero, the slope of the tangent line at any point is The tangent line will be horizontal if m = 0, that is, if 2ax + b = 0, whence Substituting this value in (1) the corresponding value of y is mum point. To translate the axes so that O' is the new origin we set The graph of this equation is a parabola (Section 30) congruent to the graph of ax2. Its axis of symmetry is vertical, and the curve runs up or down according as a is positive or negative. 1 Then since the graph of (1), plotted on the old axes, is identical with that of (6), plotted on the new axes, we have the Theorem. The graph of the quadratic function ax2 + bx + c is a parabola, with vertical axis, which is congruent to the graph of ax2. It runs up or down according as the coefficient of x2 is positive or negative. EXAMPLE. If a ball is thrown vertically upward with a velocity of 80 feet per second, its height s, in feet, after t seconds, is given by the equation 8 = 16t2 + 80t. (7) -- Find when the ball will be highest, how high it will rise, and compare the time of rising with that of falling. Construct the graph. At the highest point the velocity of the ball is zero, and we therefore seek first the velocity at any time. Replacing t by t + At and s by s + As Passing to the limit as At approaches zero, the velocity at any time is since the limit of the average velocity As/At is the instantaneous velocity. At the highest point reached by the ball v = 0, and hence t seconds. Substituting this value in (7) the maximum value of s is = The graph of this equation plotted on the new axes is identical with that of (7) plotted on the old axes. The figure shows the graph, which is readily plotted on the new axes. It passes through the old origin since |