- 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.

- schrodinger

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- jdoe0001

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

- anonymous

I dont understand what to do

- jdoe0001

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

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## More answers

- anonymous

Yeah

- 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

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

t=1 and -5?

- 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

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

- anonymous

can you help with me part b?

- jdoe0001

what's part b?

- 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

do you know what a perfect square trinomial is?

- anonymous

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

- anonymous

- jdoe0001

yes site is a bit sluggish :/

- 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

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

- 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

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

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

- anonymous

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

- jdoe0001

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

- anonymous

So is it 2?

- 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

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

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

- 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

What is a vertex?

- jdoe0001

a vertex is where the graph makes a U-turn

- anonymous

is it like parabola

- 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

what is the negative 2 for?

- jdoe0001

the x-axis \(\Large \begin{array}{ccccc}
(&-2&,&144&)\\
&x&&y
\end{array}
\)
http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiItMTZ4XjItNjR4KzgwIiwiY29sb3IiOiIjRTMxMDEwIn0seyJ0eXBlIjoxMDAwLCJ3aW5kb3ciOlsiLTE3LjEzNTU0NDM4OTAyMDYzNyIsIjIyLjUzNzMwNzE3MzQ3OTMwNiIsIjEyMy42MzAzNzA3NzAxMTUyNiIsIjE0OC4wNDQ0MzMyNzAxMTUyIl19XQ--

- anonymous

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

- 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

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

- anonymous

Just divide the graph in half right?

- jdoe0001

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

- 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

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

Thats it?

- jdoe0001

yeap

- anonymous

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

- jdoe0001

yw

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