An expression is shown below:
f(x) = -16x2 + 22x + 3
Part A: What are the x-intercepts of the graph of the f(x)? Show your work. (2 points)
Part B: Is the vertex of the graph of f(x) going to be a maximum or minimum? What are the coordinates of the vertex? Justify your answers and show your work. (3 points)
Part C: What are the steps you would use to graph f(x)? Justify that you can use the answers obtained in Part A and Part B to draw the graph. (5 points)

- anonymous

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- anonymous

@UsukiDoll this is the last one im having trouble with, can you help me? please

- Michele_Laino

part A:
we have to solve the subsequent quadratic equation:
\[\Large - 16{x^2} + 22x + 3 = 0\]

- anonymous

can you help me with that? again im new to this...

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## More answers

- Michele_Laino

I rewrite that equation as below:
\[\Large 16{x^2} - 22x - 3 = 0\]

- Michele_Laino

Next I compare that equation with this more general equation:
\[\Large a{x^2} + bx + c = 0\]
so we can write:
a=16, b=-22, and c=-3
am I right?

- anonymous

yes

- Michele_Laino

now, in order to find the requested x-intercepts, you have to apply this formula:
\[\Large x = \frac{{ - b \pm \sqrt {{b^2} - 4 \times a \times c} }}{{2 \times a}}\]
what do you get?

- anonymous

completely lost

- Michele_Laino

you have to substitute the values of a, b, and c into that formula:
\[\Large \begin{gathered}
x = \frac{{ - b \pm \sqrt {{b^2} - 4 \times a \times c} }}{{2 \times a}} = \hfill \\
\hfill \\
= \frac{{ - \left( { - 22} \right) \pm \sqrt {{{\left( { - 22} \right)}^2} - 4 \times 16 \times \left( { - 3} \right)} }}{{2 \times 16}} = ...? \hfill \\
\end{gathered} \]

- anonymous

i dont have a calculator at the moment, its hard doing it in my head

- Michele_Laino

please use windows calculator

- Michele_Laino

\[\Large \begin{gathered}
x = \frac{{ - b \pm \sqrt {{b^2} - 4 \times a \times c} }}{{2 \times a}} = \hfill \\
\hfill \\
= \frac{{ - \left( { - 22} \right) \pm \sqrt {{{\left( { - 22} \right)}^2} - 4 \times 16 \times \left( { - 3} \right)} }}{{2 \times 16}} = \hfill \\
\hfill \\
= \frac{{22 \pm \sqrt {484 + 192} }}{{32}} = ...? \hfill \\
\end{gathered} \]

- anonymous

mine isnt working at the moment...

- Michele_Laino

next step:
\[\Large \begin{gathered}
x = \frac{{ - b \pm \sqrt {{b^2} - 4 \times a \times c} }}{{2 \times a}} = \hfill \\
\hfill \\
= \frac{{ - \left( { - 22} \right) \pm \sqrt {{{\left( { - 22} \right)}^2} - 4 \times 16 \times \left( { - 3} \right)} }}{{2 \times 16}} = \hfill \\
\hfill \\
= \frac{{22 \pm \sqrt {484 + 192} }}{{32}} = \hfill \\
\hfill \\
= \frac{{22 \pm 26}}{{32}} = \begin{array}{*{20}{c}}
{\frac{{22 + 26}}{{32}} = ...?} \\
{\frac{{22 - 26}}{{32}} = ...?}
\end{array} \hfill \\
\end{gathered} \]

- anonymous

48/32 and -4/32

- anonymous

leaves you with -12/32

- Michele_Laino

yes! and we can simplify them as below:
\[\Large {x_1} = \frac{{48}}{{32}} = \frac{3}{2},\quad {x_2} = - \frac{4}{{32}} = - \frac{1}{8}\]

- Michele_Laino

what is -12/32?

- anonymous

your right, i was overr thinking..

- Michele_Laino

so we got your x-intercepts

- Michele_Laino

Now, part B:
the vertex of a parabola is the point at the top or the point at the bottom of a parabola

- Michele_Laino

we have these subsequent cases:
|dw:1439456183056:dw|

- Michele_Laino

now, in our original equation:
\[\Large f\left( x \right) = - 16{x^2} + 22x + 3\]
is a= -16, so what is your parabola?

- anonymous

lost

- Michele_Laino

the coefficient of the second degree monomial od f(x) is -16, right?

- anonymous

yes

- Michele_Laino

in that case, being -16 < 0, your parabola is concave down, namely your parabola is the one at the left place in my drawing
|dw:1439456653946:dw|

- anonymous

okay...

- Michele_Laino

now, the coordinate of the vertex, are given by the subsequent formulas:
\[\Large {x_V} = - \frac{B}{{2A}},\quad {y_V} = - \frac{{{B^2} - 4AC}}{{4A}}\]
where A= -16, B=22, and C=3

- Michele_Laino

what do you get?

- Michele_Laino

hint:
\[\Large {x_V} = - \frac{B}{{2A}} = - \frac{{22}}{{2 \times \left( { - 16} \right)}} = ...?\]

- anonymous

-22/256 =22/-32

- Michele_Laino

I got this:
\[\Large {x_V} = \frac{{22}}{{32}} = \frac{{11}}{{16}}\]

- anonymous

thank you

- Michele_Laino

now, we have to compute the y-coordinate, so we have:
\[\Large \begin{gathered}
{y_V} = - \frac{{{B^2} - 4AC}}{{4A}} = - \frac{{{{22}^2} - 4 \times \left( { - 16} \right) \times 3}}{{4 \times \left( { - 16} \right)}} = \hfill \\
\hfill \\
= - \frac{{484 + 192}}{{ - 64}} = ...? \hfill \\
\end{gathered} \]

- anonymous

my computer is acting up and this is all confusing to me, its my last question and im stressing

- anonymous

676/-64

- Michele_Laino

more precisely, we get this:
\[\Large {y_V} = \frac{{676}}{{64}} = \frac{{169}}{{16}}\]

- Michele_Laino

so the vertex V, of your parabola is the subsequent point:
\[\Large V = \left( {\frac{{11}}{{16}},\frac{{169}}{{16}}} \right)\]

- anonymous

thank you

- Michele_Laino

please wait, we have to solve part C

- anonymous

yes, im understanding everything, just upset that my computer isnt letting me use my cal.

- Michele_Laino

in order to draw your parabola, it is suffice to draw the x-intercepts and the vertex V into a xy-plane, like this:
|dw:1439457756334:dw|

- anonymous

yes

- Michele_Laino

then you have to draw a curve like this:
|dw:1439457886482:dw|

- anonymous

i understand so far

- Michele_Laino

:)

- Michele_Laino

that's all!

- anonymous

thank you!

- Michele_Laino

:)

- anonymous

trying to put everything together in sentences for part c... sadly i cant send in my drawling

- anonymous

@Michele_Laino

- Michele_Laino

we can write this:
step #1 I draw the x-intercepts, and the vertex of my parabola,
step#2 I connect those three points each other using a continuous line.

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