EXAMPLE 6 Find the points on the curve y = x - 6x2 + 7 where the tangent line is horizontal. LAL SOLUTION Horizontal tangents occur where the derivative is zero. We have 6- dx dx 4x - 12x = 4x x- - 3 Video Example.() Thus dy/dx = 0 if x = 0 or 2 - 3 = 0, that is, x = $ V3 . So the given curve has horizontal tangents when X = 0, - V3,V3 (enter your answers as a comma-separated list). The corresponding points are then the following. (x, y) = (smallest x-value) (x, y) = (x, y) = (largest x-value) XFind equations of the tangent line and normal line to the curve at the given point. (4 + 4ex, (0, 4) tangent line V = normal line VE\fThe equation of motion of a particle is s = to - 3t, where s is in meters and t is in seconds. (Assume t 2 0.) (a) Find the velocity and acceleration as functions of t. v(t) (b) Find the acceleration after 3 s. m/=2 (c) Find the acceleration when the velocity is 0. m/s2Find the points on the curve y = 2x + 3x- - 12x + 6 where the tangent line is horizontal. (x, y) = (smaller x-value) X (x, y) = (larger x-value)At what point on the curve y = 2 + 2e* - 3x is the tangent line parallel to the line 3x - y = 5? (x, y) = Illustrate by graphing the curve and both lines. y 2 -3 -2 -1 1 2 3 4 -3 -2 1 2 3 4 -21 -4 O O 8 6 2 X -4 -3 -2 -1 3 -4 -3 -2 -1 1 2 3 - 2 -4 O OEXAMPLE 3 Find the equations of the tangent line and normal line to the curve y = xv x at the point (1, 1). Illustrate the curve and thes lines. SOLUTION The derivative of R(x) = xv x = xx1/2 = x3/2 is 3/2 - 1) 3 F'(x) = 3/2 3/2 1/2 2 - normal So the slope of the tangent line at (1, 1) is f'(1) = 3/2 . Therefore an equation of the tangent line is y - 1 = 3/2 tangent x - 1 or 0.0 0.5 1.0 1.5 2.0 2.5 3.0 Video Example () X The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal of X , that is, -2/3 . Thus the equation of the normal line is