## anonymous one year ago A sandbag was thrown downward from a building. The function f(t) = -16t2 - 64t + 80 shows the height f(t), in feet, of the sandbag after t seconds: Part A: Factor the function f(t) and use the factors to interpret the meaning of the x-intercept of the function.

1. anonymous

so... .what did you get for the function factors?

2. anonymous

I dont understand what to do

3. anonymous

well, have you covered quadratic factoring yet? as in factoring quadratic equations, or equations of 2nd degree

4. anonymous

Yeah

5. anonymous

so... the x-intercepts or "zeros" or solutions of the quadratic are when y = 0, or f(x) is 0 so $$\bf f(t)=-16t^2-64t+80\implies 0=-16t^2-64t+80 \\ \quad \\ 16t^2+64t-80=0\implies 16(t^2+4t-5)=0$$ so... any idea on the factors?

6. anonymous

Im not exactly sure, im not exactly good at this. Can you explain what to do? I dont understand how to factor out 16(t^2+4t-5)=0.

7. anonymous

t=1 and -5?

8. anonymous

$$\bf f(t)=-16t^2-64t+80\implies 0=-16t^2-64t+80 \\ \quad \\ 16t^2+64t-80=0\implies 16(t^2+4t-5)=0 \\ \quad \\ t^2+4t-5=\cfrac{0}{16}\implies t^2+4t-5=0$$ how about now?

9. anonymous

yeap.... 1 and -5 are the x-intercepts

10. anonymous

can you help with me part b?

11. anonymous

what's part b?

12. anonymous

Part B: Complete the square of the expression for f(x) to determine the vertex of the graph f(x). would this be maximum or minimum on the graph?

13. anonymous

do you know what a perfect square trinomial is?

14. anonymous

Isnt the middle number supposed to be 2 times more then ones on the side>?

15. anonymous

@jdoe0001

16. anonymous

yes site is a bit sluggish :/

17. anonymous

yeap so... let us do some grouping first $$\bf f(t)=-16t^2-64t+80\implies 16t^2+64t-80=f(t) \\ \quad \\ (16t^2+64t)-80=f(t)\implies (16t^2+64t+{\color{red}{ \square }}^2)-80=f(t)$$ any ideas on what our missing number is there, to get a "perfect square trinomial" from the parenthesized group?

18. anonymous

8? OR 10? I dont know. im confused again.

19. anonymous

well... lemme simplify it some $$\bf f(t)=-16t^2-64t+80\implies 16t^2+64t-80=f(t) \\ \quad \\ (16t^2+64t)-80=f(t)\implies 16(t^2+4t)-80=f(t) \\ \quad \\ 16(t^2+4t+{\color{red}{ \square }}^2)-80=f(t)$$

20. anonymous

hmm actually, should be negative f(t) btw =) so $$\bf f(t)=-16t^2-64t+80\implies 16t^2+64t-80=-f(t) \\ \quad \\ (16t^2+64t)-80=-f(t)\implies 16(t^2+4t)-80=-f(t) \\ \quad \\ 16(t^2+4t+{\color{red}{ \square }}^2)-80=-f(t)$$

21. freckles

$ax^2+bx+c \\ =ax^2+\frac{a}{a}bx+c \\ \text{ I multiplied the second term by } \frac{a}{a} \\ \text{ I did this because I thought it would be easier } \\ \text{ for you to understand how I factor out } a \\ \text{ from the first two terms in the following step } \\ =a(x^2+\frac{1}{a}bx)+c \\ \\ =a(x^2+\frac{b}{a}x)+c \\ \text{ I'm going to leave a space } \\ \text{ this space will be meant for the number we need to add in } \\ \text{ so that we can complete the square } \\ \text{ but remember whatever I add in } \\ \text{ I have to subtract it out }$ $=\color{red}{a}(x^2+\frac{b}{a}x+\color{red}{(\frac{b}{2a})^2})+c \color{red}{-a(\frac{b}{2a})^2 }\\ =a(x+\frac{b}{2a})^2+c-a(\frac{b}{2a})^2$

22. anonymous

why is the f(t) negative? and is it 2?

23. anonymous

ohh, because I moved over to the right-hand side =)

24. anonymous

So is it 2?

25. anonymous

yes....but, notice, there's a 16 outside the group so, if we expand it, it'd be 16 * 2, or 32 so hold the mayo

26. anonymous

keep in mind, as freckles said, all we're doing is "borrowing" from our good friend Mr Zero, 0 so if we ADD $$16*2$$, we also have to SUBTRACT $$16*2^2$$ one sec

27. anonymous

how do you vind the vertex? or the maximum or minimum? Im getting lost

28. anonymous

$$\bf 16(t^2+4t+{\color{red}{ 2 }}^2-{\color{red}{ 2}}^2)-80=-f(t) \\ \quad \\ 16(t^2+4t+{\color{red}{ 2}}^2)-(16\cdot {\color{red}{ 2}}^2)-80=-f(t) \\ \quad \\ 16(t+2)^2-64-80=-f(t) \\ \quad \\ f(t)=-16(t+2)^2+64+80 \\ \quad \\ f(t)=-16(t+2)^2+144\implies f(t)=-16(t-{\color{brown}{ (-2)}})^2+{\color{blue}{ 144}} \\ \quad \\ \quad \\ y=(x-{\color{brown}{ h}})^2+{\color{blue}{ k}}\\ x=(y-{\color{blue}{ k}})^2+{\color{brown}{ h}}\qquad\qquad vertex\ ({\color{brown}{ h}},{\color{blue}{ k}})$$ see the vertex now?

29. anonymous

What is a vertex?

30. anonymous

a vertex is where the graph makes a U-turn

31. anonymous

is it like parabola

32. anonymous

also, notice, the number in front of the parentheses, is negative, -16 that means, the parabola opens downards, or it is going down|dw:1443656451839:dw|

33. anonymous

what is the negative 2 for?

34. anonymous

so the highest point is 144? and it would have a maximum not a minimum?

35. anonymous

is a maximum because the parabola goes up up up up, reaches there, makes a U-turn, then goes back down down down to infinity so the "maximum" point it reaches, is at the vertex, at -2, 144

36. anonymous

Part C: Use your answer from part B to determine the axis symmerty

37. anonymous

Just divide the graph in half right?

38. anonymous

axis of symmetry ...yes, divide it in half|dw:1443656899492:dw|

39. anonymous

So how do describe that? with just a number? or im confused. i know you have to go down the middle but whats after that?

40. anonymous

the axis of symmetry is just a line thus you write the equation of that line in this case is a vertical line, which is $$\bf x= -2$$

41. anonymous

Thats it?

42. anonymous

yeap

43. anonymous

Awe, thank you so much for your help! I have such a better understanding now!

44. anonymous

yw