anonymous
  • anonymous
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)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@UsukiDoll this is the last one im having trouble with, can you help me? please
Michele_Laino
  • Michele_Laino
part A: we have to solve the subsequent quadratic equation: \[\Large - 16{x^2} + 22x + 3 = 0\]
anonymous
  • anonymous
can you help me with that? again im new to this...

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Michele_Laino
  • Michele_Laino
I rewrite that equation as below: \[\Large 16{x^2} - 22x - 3 = 0\]
Michele_Laino
  • 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
  • anonymous
yes
Michele_Laino
  • 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
  • anonymous
completely lost
Michele_Laino
  • 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
  • anonymous
i dont have a calculator at the moment, its hard doing it in my head
Michele_Laino
  • Michele_Laino
please use windows calculator
Michele_Laino
  • 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
  • anonymous
mine isnt working at the moment...
Michele_Laino
  • 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
  • anonymous
48/32 and -4/32
anonymous
  • anonymous
leaves you with -12/32
Michele_Laino
  • 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
  • Michele_Laino
what is -12/32?
anonymous
  • anonymous
your right, i was overr thinking..
Michele_Laino
  • Michele_Laino
so we got your x-intercepts
Michele_Laino
  • 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
  • Michele_Laino
we have these subsequent cases: |dw:1439456183056:dw|
Michele_Laino
  • 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
  • anonymous
lost
Michele_Laino
  • Michele_Laino
the coefficient of the second degree monomial od f(x) is -16, right?
anonymous
  • anonymous
yes
Michele_Laino
  • 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
  • anonymous
okay...
Michele_Laino
  • 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
  • Michele_Laino
what do you get?
Michele_Laino
  • Michele_Laino
hint: \[\Large {x_V} = - \frac{B}{{2A}} = - \frac{{22}}{{2 \times \left( { - 16} \right)}} = ...?\]
anonymous
  • anonymous
-22/256 =22/-32
Michele_Laino
  • Michele_Laino
I got this: \[\Large {x_V} = \frac{{22}}{{32}} = \frac{{11}}{{16}}\]
anonymous
  • anonymous
thank you
Michele_Laino
  • 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
  • anonymous
my computer is acting up and this is all confusing to me, its my last question and im stressing
anonymous
  • anonymous
676/-64
Michele_Laino
  • Michele_Laino
more precisely, we get this: \[\Large {y_V} = \frac{{676}}{{64}} = \frac{{169}}{{16}}\]
Michele_Laino
  • Michele_Laino
so the vertex V, of your parabola is the subsequent point: \[\Large V = \left( {\frac{{11}}{{16}},\frac{{169}}{{16}}} \right)\]
anonymous
  • anonymous
thank you
Michele_Laino
  • Michele_Laino
please wait, we have to solve part C
anonymous
  • anonymous
yes, im understanding everything, just upset that my computer isnt letting me use my cal.
Michele_Laino
  • 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
  • anonymous
yes
Michele_Laino
  • Michele_Laino
then you have to draw a curve like this: |dw:1439457886482:dw|
anonymous
  • anonymous
i understand so far
Michele_Laino
  • Michele_Laino
:)
Michele_Laino
  • Michele_Laino
that's all!
anonymous
  • anonymous
thank you!
Michele_Laino
  • Michele_Laino
:)
anonymous
  • anonymous
trying to put everything together in sentences for part c... sadly i cant send in my drawling
anonymous
  • anonymous
@Michele_Laino
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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.