anonymous
  • anonymous
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.
Algebra
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
jdoe0001
  • jdoe0001
so... .what did you get for the function factors?
anonymous
  • anonymous
I dont understand what to do
jdoe0001
  • jdoe0001
well, have you covered quadratic factoring yet? as in factoring quadratic equations, or equations of 2nd degree

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anonymous
  • anonymous
Yeah
jdoe0001
  • jdoe0001
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?
anonymous
  • 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.
anonymous
  • anonymous
t=1 and -5?
jdoe0001
  • jdoe0001
\(\bf f(t)=-16t^2-64t+80\implies 0=-16t^2-64t+80 \\ \quad \\ 16t^2+64t-80=0\implies 16(t^2+4t-5)=0 \\ \quad \\ t^2+4t-5=\cfrac{0}{16}\implies t^2+4t-5=0\) how about now?
jdoe0001
  • jdoe0001
yeap.... 1 and -5 are the x-intercepts
anonymous
  • anonymous
can you help with me part b?
jdoe0001
  • jdoe0001
what's part b?
anonymous
  • 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?
jdoe0001
  • jdoe0001
do you know what a perfect square trinomial is?
anonymous
  • anonymous
Isnt the middle number supposed to be 2 times more then ones on the side>?
anonymous
  • anonymous
@jdoe0001
jdoe0001
  • jdoe0001
yes site is a bit sluggish :/
jdoe0001
  • jdoe0001
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?
anonymous
  • anonymous
8? OR 10? I dont know. im confused again.
jdoe0001
  • jdoe0001
well... lemme simplify it some \(\bf f(t)=-16t^2-64t+80\implies 16t^2+64t-80=f(t) \\ \quad \\ (16t^2+64t)-80=f(t)\implies 16(t^2+4t)-80=f(t) \\ \quad \\ 16(t^2+4t+{\color{red}{ \square }}^2)-80=f(t)\)
jdoe0001
  • jdoe0001
hmm actually, should be negative f(t) btw =) so \(\bf f(t)=-16t^2-64t+80\implies 16t^2+64t-80=-f(t) \\ \quad \\ (16t^2+64t)-80=-f(t)\implies 16(t^2+4t)-80=-f(t) \\ \quad \\ 16(t^2+4t+{\color{red}{ \square }}^2)-80=-f(t)\)
freckles
  • freckles
\[ax^2+bx+c \\ =ax^2+\frac{a}{a}bx+c \\ \text{ I multiplied the second term by } \frac{a}{a} \\ \text{ I did this because I thought it would be easier } \\ \text{ for you to understand how I factor out } a \\ \text{ from the first two terms in the following step } \\ =a(x^2+\frac{1}{a}bx)+c \\ \\ =a(x^2+\frac{b}{a}x)+c \\ \text{ I'm going to leave a space } \\ \text{ this space will be meant for the number we need to add in } \\ \text{ so that we can complete the square } \\ \text{ but remember whatever I add in } \\ \text{ I have to subtract it out } \] \[=\color{red}{a}(x^2+\frac{b}{a}x+\color{red}{(\frac{b}{2a})^2})+c \color{red}{-a(\frac{b}{2a})^2 }\\ =a(x+\frac{b}{2a})^2+c-a(\frac{b}{2a})^2\]
anonymous
  • anonymous
why is the f(t) negative? and is it 2?
jdoe0001
  • jdoe0001
ohh, because I moved over to the right-hand side =)
anonymous
  • anonymous
So is it 2?
jdoe0001
  • jdoe0001
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
jdoe0001
  • jdoe0001
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
anonymous
  • anonymous
how do you vind the vertex? or the maximum or minimum? Im getting lost
jdoe0001
  • jdoe0001
\(\bf 16(t^2+4t+{\color{red}{ 2 }}^2-{\color{red}{ 2}}^2)-80=-f(t) \\ \quad \\ 16(t^2+4t+{\color{red}{ 2}}^2)-(16\cdot {\color{red}{ 2}}^2)-80=-f(t) \\ \quad \\ 16(t+2)^2-64-80=-f(t) \\ \quad \\ f(t)=-16(t+2)^2+64+80 \\ \quad \\ f(t)=-16(t+2)^2+144\implies f(t)=-16(t-{\color{brown}{ (-2)}})^2+{\color{blue}{ 144}} \\ \quad \\ \quad \\ y=(x-{\color{brown}{ h}})^2+{\color{blue}{ k}}\\ x=(y-{\color{blue}{ k}})^2+{\color{brown}{ h}}\qquad\qquad vertex\ ({\color{brown}{ h}},{\color{blue}{ k}})\) see the vertex now?
anonymous
  • anonymous
What is a vertex?
jdoe0001
  • jdoe0001
a vertex is where the graph makes a U-turn
anonymous
  • anonymous
is it like parabola
jdoe0001
  • jdoe0001
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|
anonymous
  • anonymous
what is the negative 2 for?
jdoe0001
  • jdoe0001
the x-axis \(\Large \begin{array}{ccccc} (&-2&,&144&)\\ &x&&y \end{array} \) http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiItMTZ4XjItNjR4KzgwIiwiY29sb3IiOiIjRTMxMDEwIn0seyJ0eXBlIjoxMDAwLCJ3aW5kb3ciOlsiLTE3LjEzNTU0NDM4OTAyMDYzNyIsIjIyLjUzNzMwNzE3MzQ3OTMwNiIsIjEyMy42MzAzNzA3NzAxMTUyNiIsIjE0OC4wNDQ0MzMyNzAxMTUyIl19XQ--
anonymous
  • anonymous
so the highest point is 144? and it would have a maximum not a minimum?
jdoe0001
  • jdoe0001
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
anonymous
  • anonymous
Part C: Use your answer from part B to determine the axis symmerty
anonymous
  • anonymous
Just divide the graph in half right?
jdoe0001
  • jdoe0001
axis of symmetry ...yes, divide it in half|dw:1443656899492:dw|
anonymous
  • 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?
jdoe0001
  • jdoe0001
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 \(\bf x= -2\)
anonymous
  • anonymous
Thats it?
jdoe0001
  • jdoe0001
yeap
anonymous
  • anonymous
Awe, thank you so much for your help! I have such a better understanding now!
jdoe0001
  • jdoe0001
yw

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