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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.

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  1. jdoe0001
    • one year ago
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    so... .what did you get for the function factors?

  2. anonymous
    • one year ago
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    I dont understand what to do

  3. jdoe0001
    • one year ago
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    well, have you covered quadratic factoring yet? as in factoring quadratic equations, or equations of 2nd degree

  4. anonymous
    • one year ago
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    Yeah

  5. jdoe0001
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    t=1 and -5?

  8. jdoe0001
    • one year ago
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    \(\bf f(t)=-16t^2-64t+80\implies 0=-16t^2-64t+80 \\ \quad \\ 16t^2+64t-80=0\implies 16(t^2+4t-5)=0 \\ \quad \\ t^2+4t-5=\cfrac{0}{16}\implies t^2+4t-5=0\) how about now?

  9. jdoe0001
    • one year ago
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    yeap.... 1 and -5 are the x-intercepts

  10. anonymous
    • one year ago
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    can you help with me part b?

  11. jdoe0001
    • one year ago
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    what's part b?

  12. anonymous
    • one year ago
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    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. jdoe0001
    • one year ago
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    do you know what a perfect square trinomial is?

  14. anonymous
    • one year ago
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    Isnt the middle number supposed to be 2 times more then ones on the side>?

  15. anonymous
    • one year ago
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    @jdoe0001

  16. jdoe0001
    • one year ago
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    yes site is a bit sluggish :/

  17. jdoe0001
    • one year ago
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    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
    • one year ago
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    8? OR 10? I dont know. im confused again.

  19. jdoe0001
    • one year ago
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    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. jdoe0001
    • one year ago
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    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
    • one year ago
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    \[ax^2+bx+c \\ =ax^2+\frac{a}{a}bx+c \\ \text{ I multiplied the second term by } \frac{a}{a} \\ \text{ I did this because I thought it would be easier } \\ \text{ for you to understand how I factor out } a \\ \text{ from the first two terms in the following step } \\ =a(x^2+\frac{1}{a}bx)+c \\ \\ =a(x^2+\frac{b}{a}x)+c \\ \text{ I'm going to leave a space } \\ \text{ this space will be meant for the number we need to add in } \\ \text{ so that we can complete the square } \\ \text{ but remember whatever I add in } \\ \text{ I have to subtract it out } \] \[=\color{red}{a}(x^2+\frac{b}{a}x+\color{red}{(\frac{b}{2a})^2})+c \color{red}{-a(\frac{b}{2a})^2 }\\ =a(x+\frac{b}{2a})^2+c-a(\frac{b}{2a})^2\]

  22. anonymous
    • one year ago
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    why is the f(t) negative? and is it 2?

  23. jdoe0001
    • one year ago
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    ohh, because I moved over to the right-hand side =)

  24. anonymous
    • one year ago
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    So is it 2?

  25. jdoe0001
    • one year ago
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    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. jdoe0001
    • one year ago
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    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
    • one year ago
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    how do you vind the vertex? or the maximum or minimum? Im getting lost

  28. jdoe0001
    • one year ago
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    \(\bf 16(t^2+4t+{\color{red}{ 2 }}^2-{\color{red}{ 2}}^2)-80=-f(t) \\ \quad \\ 16(t^2+4t+{\color{red}{ 2}}^2)-(16\cdot {\color{red}{ 2}}^2)-80=-f(t) \\ \quad \\ 16(t+2)^2-64-80=-f(t) \\ \quad \\ f(t)=-16(t+2)^2+64+80 \\ \quad \\ f(t)=-16(t+2)^2+144\implies f(t)=-16(t-{\color{brown}{ (-2)}})^2+{\color{blue}{ 144}} \\ \quad \\ \quad \\ y=(x-{\color{brown}{ h}})^2+{\color{blue}{ k}}\\ x=(y-{\color{blue}{ k}})^2+{\color{brown}{ h}}\qquad\qquad vertex\ ({\color{brown}{ h}},{\color{blue}{ k}})\) see the vertex now?

  29. anonymous
    • one year ago
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    What is a vertex?

  30. jdoe0001
    • one year ago
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    a vertex is where the graph makes a U-turn

  31. anonymous
    • one year ago
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    is it like parabola

  32. jdoe0001
    • one year ago
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    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
    • one year ago
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    what is the negative 2 for?

  34. anonymous
    • one year ago
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    so the highest point is 144? and it would have a maximum not a minimum?

  35. jdoe0001
    • one year ago
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    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
    • one year ago
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    Part C: Use your answer from part B to determine the axis symmerty

  37. anonymous
    • one year ago
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    Just divide the graph in half right?

  38. jdoe0001
    • one year ago
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    axis of symmetry ...yes, divide it in half|dw:1443656899492:dw|

  39. anonymous
    • one year ago
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    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. jdoe0001
    • one year ago
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    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
    • one year ago
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    Thats it?

  42. jdoe0001
    • one year ago
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    yeap

  43. anonymous
    • one year ago
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    Awe, thank you so much for your help! I have such a better understanding now!

  44. jdoe0001
    • one year ago
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    yw

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