6.2 6.3: Finding x-intercepts

1. 6.2 & 6.3:Finding x-intercepts Finding by graphing Finding using factoring Related word problems

2. 6.2 � Finding x-intercepts by graphing In addition to the vertex and y-intercept, we are often interested in finding the points of the parabola representing the x-intercepts One way to find x-intercepts is by graphing � make a table of values, plot the points, and try to estimate the coordinates of the x-intercepts Graphing is not a particularly efficient way of finding intercepts � you may need a LARGE table, and even then you may still end up having to estimate the coordinates if they do not occur at �nice � numbers A parabola may have either 0, 1, or 2 x-intercepts (see the examples on the next slide)

3. 3 possibilities:

4. Example 2-1a

5. Example 2-1a

6. Example 2-1a

7. 6.2 Solving by factoring In this section we discuss how to find x-intercepts algebraically by factoring This technique will work if the x-intercepts occur at rational numbers (3, -8, � , 2 � , etc.) To find irrational x-intercepts we will need to use other techniques (we will discuss them after the break)

8. Terminology An equation of the form: ax2 + bx + c = 0 is called a quadratic equation The solutions to quadratic equations are called the ROOTS of the equation Since x-intercepts have y-coordinates of 0, for a quadratic function f(x) = ax2 + bx + c, we can find the x-intercepts by finding the roots of the quadratic equation: ax2 + bx + c = 0 Complicating things further, we often called x-intercepts for a quadratic function the ZEROES of the function (zeroes and x-intercepts are synonymous)

9. Solving quadratic equations by factoring Some, but not all, quadratic equations can be solved by factoring Specifically, if the roots are rational ,we can find them by factoring To find roots, we use the ZERO PRODUCT PROPERTY This property states: If a*b = 0, then a = 0, b = 0, or both a and b = 0 Basically we set up a quadratic equation so that one side equals 0, factor the quadratic, then set each factor equal to 0 and solve!

10. Example 3-1a

11. Example 3-2a

12. Example 3-1a

13. Example 3-1a

14. Example 3-1b

15. Example 3-2b

16. Writing a quadratic equation, given its roots If (x � A) is a factor of a quadratic equation, then A is a root of the equation Thus each root generates a factor, and the product of these factors should be zero So if you have roots at 3 and 7, then the corresponding quadratic equation is: (x � 3)(x � 7) = 0 Now multiply it out! X2 � 10x + 21 = 0

17. However� Usually it is required that the a,b, and c coefficients of the quadratic equation are integers If our roots are fractions, this is problematic We can make a,b, and c integers by multiplying through by an appropriate number This is OK since one side of the equation is 0, and multiplying by a constant will not change that

18. Example 3-4a

19. Example 3-4a

20. Example 3-4b

21. Related word problems In this chapter, we often look at problems involving objects that are falling due to gravity, or are launched upwards and then fall back down due to gravity For these problems we use the formula: H(t) = -16t2 + V0t + H0 This function gives the height of the object above the ground, t seconds after it is dropped/launched Where V0 is the starting speed (for launched object) and H0 is the starting height

22. Two types of problems involving this formula If you are asked to find the maximum height a launched object will reach, find the vertex of this parabola If you are asked to find the time when the object will reach a certain height, set h(t) equal to that height, then solve the quadratic equation

23. Example Sammie is launched upwards from a trampoline at a speed of 26 ft/sec, and the trampoline sits atop a 35 foot building. Use the formula H(t) = -16t2 + 26t + 35 to answer each question A) What is the maximum height above the ground that Sammie will reach? B) At what time will Sammie land back on the ground below the building (assume we have a safety landing pad)

24. Example Sammie is launched upwards from a trampoline at a speed of 26 ft/sec, and the trampoline sits atop a 35 foot building. Use the formula H(t) = -16t2 + 26t + 35 to answer each question A) What is the maximum height above the ground that Sammie will reach? Find the vertex: a = -16 and b = 26 So t = -b/2a = -(-26) / (2 * 16) = 26/32 = .8125 seconds Max height = H(.308) = -16 (.8125)2 + 26 (.8125) + 35 = 45.56 feet!

25. Example Sammie is launched upwards from a trampoline at a speed of 26 ft/sec, and the trampoline sits atop a 35 foot building. Use the formula H(t) = -16t2 + 26t + 35 to answer each question B) At what time will Sammie land back on the ground below the building (assume we have a safety landing pad) Set H(t) = 0 and solve the quadratic equation 0 = -16t2 + 26t + 35 0 = -1 (16t2 � 26t � 35) 0 = -1 (8t + 7)(2t � 5) So 8t + 7 = 0 and 2t � 5 = 0 This gives solutions of -7/8 sec and 2.5 sec However -7/8 sec is not realistic for this problem so Sammie will land on the ground 2.5 seconds after launch!!

26. Another word problem For a cube, if a side measures x units, then each side has an area of x2 Since the cube has 6 sides of equal area, the SURFACE AREA = 6x2 Suppose I have a cube, and I decrease each side by 3 inches. If the new surface area of the cube is 433.5 in2, what did each side measure originally? The new side = x � 3 Since SA = 6 (side)2, 433.5 = 6(x � 3)2 72.25 = (x � 3)2 Now square root each side! � 8.5 = x � 3 x = 3 � 8.5 = 11.5 or � 5.5 Since a side can�t have a negative measure, the cube originally had sides measuring 11.5 inches!

27. HOMEWORK Pg. 304: 14 � 44 evens