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Polynomial Functions

Polynomial Functions

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Polynomial Functions

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  1. Polynomial Functions

  2. Polynomial Functions • An equation having a variable raised to an exponent of a nonnegative integer • Degree: the highest power on the variable • i.e. f(x) = x2 + 1 f(x) = x3 + 1 f(x) = x+ 2 f(x) = 3 second degree third degree first degree zero degree (no variable)

  3. Zero Degree Polynomials Constant Functions

  4. Constant Functions • A constant function is described by the rule f(x) = b where b can be any number • The rate of change (slope) is zero • Horizontalline with equation y = b (0, 3) y = 3

  5. First Degree Polynomials Linear Functions

  6. Linear Functions • A linear function is described by the rule f(x) = ax + b where a cannot equal 0 • The parameter “b” is the y-intercept • The parameter “a” represents the rate of change (slope) • If a > 0, rate of change is positive • If a < 0, rate of change is negative Rate of Change = y2 – y1 x2 – x1Point 1 (x1, y1)Point 2 ( x2, y2) 2 -2 2

  7. Second Degree Polynomials Quadratic Functions

  8. Quadratic Function a < 0 a > 0 Standard Formf(x) = a(x-h)2 + k Vertex: (h,k)Axis of symmetry: x = hZeros: h Factored Formf(x) = a(x-x1)(x-x2) Zeros: x1 and x2Axis of Symmetry: x = General Formf(x) = ax2 + bx + c Vertex: Axis of symmetry: x = Zeros: y intercept: c

  9. Example The following formula links the speed v in km/h and the distance d in metres necessary to bring a car to a complete stop in case of emergency. 100d = v2 + 20v At what speed is a car travelling if it needs 63 m to come to a complete stop? 100(63) = v2 + 20v 6300 = v2 + 20v 0 = v2 + 20v - 6300

  10. Example f(x) = a(x-h)2 + k The parabola represented below crosses the x-axis at the points (-1, 0) and (3, 0) and its vertex is the point P(1, ‑4). What is the equation of the parabola graphed above in general form? f(x) = a(x-1)2 - 4 0= a(3-1)2 - 4 0= 4a- 4 a=1 f(x) = (x-1)2 - 4 f(x) = (x-1)(x-1) - 4 f(x) = x2- x – x +1 - 4 y = x2 2x 3

  11. Example • One leg of a right triangle exceeds the other leg by four inches. The hypotenuse is 20 inches. Find the length of the longer leg of the right triangle. c2 = a2 + b2 20 202 = x2 + (x + 4) 2 x 400 = x2 + (x + 4)(x + 4) 400 = x2 + (x2 + 8x + 16) 0 = 2x2 + 8x -384 x + 4

  12. Example The height, h(t), in feet of an object above the ground is given by h(t) = -16t2 + 64t + 190, where t is the time in seconds. a) Find the time it takes the object to strike the ground b) Find the maximum height of the object. Vertex:

  13. Example A diverjumps off the side of an ingroundswimming pool and reaches a depth of 2.6 metersat a distance of 1.95 m awayfrom the ledge. He enters the water at a distance of 1.76 metersfrom the ledge. Determine the maximum heightduringhistrajectoryassumingthathefollowed a parabolic motion. Vertex: f(x) = a(x-x1)(x-x2) f(x) = a(x-0)(x-1.76) -2.6 = a(1.95-0)(1.95-1.76) -2.6 = a(1.95)(0.19) -2.6 = 0.3705a a = -7.02 (1.76,0) f(x) = -7.02x(x-1.76) f(x) = -7.02x2 + 12.36x (1.95, -2.6)