Grade 10_Math-Lesson 2-3 Graphs of Polynomial Functions .pptx
- 3. Graphs Polynomial Functions The end behavior tells you how a graph looks like at each end-whether the graph rises or falls to the left or the right. To determine the end behavior of the graph of a polynomial function, you will use the Leading Coefficient Test. The graph of a polynomial function P (x) of degree n has at most n-1 turning points. the highest exponent 𝒇 𝒙 = 𝟑𝒙𝟒 + 𝟒𝒙𝟑 − 𝟑𝒙𝟐 − 𝟒𝒙 − 𝟑 The degree is 4 Turning Points: 4-1 = 3
- 4. 𝒇 𝒙 = 𝒙𝟑 + 𝟓𝒙𝟐 + 𝟑𝒙 − 𝟑 𝒇 𝒙 = 𝒙𝟒 + 𝟐𝒙𝟑 − 𝟑𝒙𝟐 − 𝟒𝒙 − 𝟑 𝒇 𝒙 = 𝒙𝟓 + 𝟑𝒙𝟒 + 𝟒𝒙𝟑 − 𝟑𝒙𝟐 − 𝟒𝒙 − 𝟑 𝒇 𝒙 = 𝟐𝒙𝟐 − 𝟑𝒙 − 𝟔 𝒇 𝒙 = 𝟓𝒙 −2 𝒇 𝒙 = 2
- 5. Behaviour of the graph of polynomial functions 𝒂𝒏 > 𝟎 𝒑𝒐𝒔𝒊𝒕𝒊𝒗𝒆 𝒂𝒏 < 𝟎 𝒏𝒆𝒈𝒂𝒕𝒊𝒗𝒆 n (degree) is even n(degree) is odd 𝑟𝑖𝑠𝑒𝑠 𝑡𝑜 𝑡ℎ𝑒 𝑙𝑒𝑓𝑡 𝑎𝑛𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡 𝑓𝑎𝑙𝑙𝑠 𝑡𝑜 𝑡ℎ𝑒 𝑙𝑒𝑓𝑡 𝑎𝑛𝑑 𝑟𝑖𝑠𝑒𝑠 𝑡𝑜 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡 𝑓𝑎𝑙𝑙𝑠 𝑡𝑜 𝑡ℎ𝑒 𝑙𝑒𝑓𝑡 𝑎𝑛𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡 𝑟𝑖𝑠𝑒𝑠 𝑡𝑜 𝑡ℎ𝑒 𝑙𝑒𝑓𝑡 𝑎𝑛𝑑 𝑓𝑎𝑙𝑙𝑠 𝑡𝑜 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡
- 6. To help you in graphing polynomial functions accurately, consider the following steps: Step 1: Determine the end behavior of the graph by using the Leading Coefficient Test Step 2: Determine the maximum number of turning points or points where the graph changes in direction. The next theorem gives the number of turning points of the graph of a polynomial function. Step 3: Find the x-intercepts or zeros of the function by setting f(x) = 0 and then solving for x. Keep in mind that when (x-r)𝑘 is a factor of the polynomial and k is even, the graph touches the x-axis at r and changes direction. When k is odd, the graph crosses the x-axis at r. Step 4: Find the y-intercept of the function by finding the value of f(x) at x=0, or solve for f(0).
- 7. To help you in graphing polynomial functions accurately, consider the following steps: Step 5: Construct a table of signs. This will help you determine the sign of the function given its factored form. The sign of the function will determine whether the graph appears below or above the x-axis. Step 6: Find other points, if needed. Sometimes, it is necessary to find points that are in between two zeros or two x-intercepts to have a more accurate graph. Step 7: Plot the points found in steps 3,4 and 6. Use the information gathered in steps 1, 2, 3, and 5 to graph the function. Always remember that the graph of a polynomial function is always a smooth curve, without any breaks.
- 8. 𝑷 𝒙 = 𝒙𝟑 + 𝟑𝒙𝟐 − 𝒙 − 𝟑 a. Leading coefficient test Degree: Odd (3) Leading Coefficient: Positive (1) Case 1: If the leading coefficient 𝒂𝒏 is positive and n is odd, the graph falls to the left and rises to the right. b. x-intercepts 𝑷 𝒙 = 𝒙𝟑 + 𝟑𝒙𝟐 − 𝒙 − 𝟑 The possible rational roots are: ±1 ; ±3 1 3 -1 -3 1 1 1 4 4 3 3 0 Since R = 0, then 1 is a root 𝑷 𝒙 = 𝒙𝟐 + 𝟒𝒙 + 𝟑 (𝒙 + 𝟑)(𝒙 + 𝟏) 𝒙 + 𝟑 = 𝟎 x= −𝟑 𝒙 + 𝟏 = 𝟎 𝒙 = −𝟏
- 9. b. y-intercepts 𝑷 𝟎 = (𝟎)𝟑 +𝟑(𝟎) 𝟐 − 𝟎 − 𝟑 (𝒙 − 𝟏)(𝒙 + 𝟑)(𝒙 + 𝟏) 𝑷 𝟎 = −𝟑 The actual roots : 1 , -3 , -1 Polynomial in factored forms (1 , 0) (-3, 0) (-1, 0) (0,-3)
- 10. 𝑷 𝒙 = 𝟐(𝒙 − 𝟓)(𝒙 + 𝟐)𝟐 𝒙 − 𝟓 = 𝟎 𝒙 = 𝟓 (𝒙 + 𝟐)𝟐 = 𝟎 (𝒙 + 𝟐)𝟐 = 𝟎 𝒙 + 𝟐 = 𝟎 𝒙 = −𝟐 a. Leading coefficient test Degree: Odd (3) Leading Coefficient: Positive (2) b. x-intercepts b. y-intercepts 𝑷 𝟎 = 𝟐(𝟎 − 𝟓)(𝟎 + 𝟐)𝟐 𝑷 𝟎 = −𝟏𝟎(𝟒) 𝑷 𝟎 = −𝟒𝟎
- 11. 𝑷 𝒙 = 𝒙𝟑 + 𝟔𝒙𝟐 + 𝟑𝒙 − 𝟏𝟎 a. Leading coefficient test Degree: Odd (3) Leading Coefficient: Positive (1) Case 1: If the leading coefficient 𝒂𝒏 is positive and n is odd, the graph falls to the left and rises to the right. b. x-intercepts 𝑷 𝒙 = 𝒙𝟑 + 𝟔𝒙𝟐 + 𝟑𝒙 − 𝟏𝟎 The possible rational roots are: ±1 ; ±2; ±5; ±10 1 6 3 -10 -5 1 -5 1 -5 -2 10 0 Since R = 0, then -5 is a root 𝑷 𝒙 = 𝒙𝟐 + 𝒙 − 𝟐 𝒙 + 𝟓 = 𝟎 x= −𝟓 𝒙 + 𝟐 = 𝟎 𝒙 = −𝟐 𝒚 = (𝒙 + 𝟓)(𝒙 + 𝟐)(𝒙 − 𝟏) 𝑷 𝒙 = (𝒙 + 𝟓)(𝒙𝟐 + 𝒙 − 𝟐) 𝒙 − 𝟏 = 𝟎 𝒙 = 𝟏 −𝟓, 𝟎 ; −𝟐, 𝟎 ; (𝟏, 𝟎)
- 12. b. y-intercepts 𝑷 𝟎 = (𝟎)𝟑 +𝟔(𝟎) 𝟐 + 𝟑(𝟎) − 𝟏𝟎 𝑷 𝟎 = −𝟏𝟎 𝑷 𝒙 = 𝒙𝟑 + 𝟔𝒙𝟐 + 𝟑𝒙 − 𝟏𝟎 −𝟓, 𝟎 ; −𝟐, 𝟎 ; 𝟏, 𝟎 ; (𝟎 , −𝟏𝟎) To get a more accurate curve, find some points between the x-intercepts x = -4 x = -1 x = 0 𝒚 = (𝒙 + 𝟓)(𝒙 + 𝟐)(𝒙 − 𝟏) 𝒚 = (𝒙 + 𝟓)(𝒙 + 𝟐)(𝒙 − 𝟏) 𝒚 = (−𝟒 + 𝟓)(−𝟒 + 𝟐)(−𝟒 − 𝟏) 𝒚 = (𝟏)(−𝟐)(−𝟓) 𝒚 = 𝟏𝟎 (−𝟒, 𝟏𝟎 𝒚 = (𝒙 + 𝟓)(𝒙 + 𝟐)(𝒙 − 𝟏) 𝒚 = (−𝟏 + 𝟓)(−𝟏 + 𝟐)(−𝟏 − 𝟏) 𝒚 = (𝟒)(𝟏)(−𝟐) 𝒚 = −𝟖 (−𝟏 , −𝟖) 𝒚 = (𝒙 + 𝟓)(𝒙 + 𝟐)(𝒙 − 𝟏) 𝒚 = (𝟎 + 𝟓)(𝟎 + 𝟐)(𝟎 − 𝟏) 𝒚 = (𝟓)(𝟐)(−𝟏) 𝒚 = −𝟏𝟎 (𝟎 , −𝟏𝟎)
- 13. Intervals x<-5 -5<x<-2 -2<x<1 x>1 Test Value x+5 x+2 x-1 y=(x+5)(x+2)(x-1) Position of the curve
- 14. 𝑷 𝒙 = 𝒙𝟑 + 𝟔𝒙𝟐 + 𝟑𝒙 − 𝟏𝟎 𝒚 = (𝒙 + 𝟓)(𝒙 + 𝟐)(𝒙 − 𝟏) −𝟓, 𝟎 ; −𝟐, 𝟎 ; 𝟏, 𝟎 ; (𝟎 , −𝟏𝟎) (−𝟏 , −𝟖) (𝟎 , −𝟏𝟎) x and y intercepts: Other points: (−𝟒 , 𝟏𝟎)
- 15. Activity 1_W1Q2 Answer Practice 2-3 Use the Leading Coefficient Test to determine the end behavior of each polynomial no. 1-15 on page 86
- 16. Behaviour of the graph of polynomial functions 𝒂𝒏 > 𝟎 𝒑𝒐𝒔𝒊𝒕𝒊𝒗𝒆 𝒂𝒏 < 𝟎 𝒏𝒆𝒈𝒂𝒕𝒊𝒗𝒆 n (degree) is even n(degree) is odd 𝑟𝑖𝑠𝑒𝑠 𝑡𝑜 𝑡ℎ𝑒 𝑙𝑒𝑓𝑡 𝑎𝑛𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡 𝑓𝑎𝑙𝑙𝑠 𝑡𝑜 𝑡ℎ𝑒 𝑙𝑒𝑓𝑡 𝑎𝑛𝑑 𝑟𝑖𝑠𝑒𝑠 𝑡𝑜 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡 𝑓𝑎𝑙𝑙𝑠 𝑡𝑜 𝑡ℎ𝑒 𝑙𝑒𝑓𝑡 𝑎𝑛𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡 𝑟𝑖𝑠𝑒𝑠 𝑡𝑜 𝑡ℎ𝑒 𝑙𝑒𝑓𝑡 𝑎𝑛𝑑 𝑓𝑎𝑙𝑙𝑠 𝑡𝑜 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡
- 17. Quiz 1 _W1Q2 Find the zero/s (x-intercept), and (y-intercept) of the following polynomial functions and graph on the Cartesian plane 𝟏. 𝒇 𝒙 = 𝒙𝟑 + 𝟒𝒙𝟐 + 𝒙 − 𝟔 𝟑. 𝒉 𝒙 = (𝟑𝒙 − 𝟐)(𝒙 + 𝟐)𝟐 𝟐. 𝒈 𝒙 = 𝒙𝟐 − 𝟒𝒙 + 𝟒 𝟒. 𝒉 𝒙 = (𝒙 + 𝟏)(𝒙 − 𝟐)(𝒙 + 𝟓) 𝟓. 𝒉 𝒙 = (𝟐 − 𝒙)(𝒙 + 𝟏)𝟐