## anonymous one year ago 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)

1. anonymous

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

2. Michele_Laino

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

3. anonymous

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

4. Michele_Laino

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

5. 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?

6. anonymous

yes

7. 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?

8. anonymous

completely lost

9. Michele_Laino

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

10. anonymous

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

11. Michele_Laino

12. Michele_Laino

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

13. anonymous

mine isnt working at the moment...

14. Michele_Laino

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

15. anonymous

48/32 and -4/32

16. anonymous

leaves you with -12/32

17. Michele_Laino

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

18. Michele_Laino

what is -12/32?

19. anonymous

your right, i was overr thinking..

20. Michele_Laino

21. 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

22. Michele_Laino

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

23. 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?

24. anonymous

lost

25. Michele_Laino

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

26. anonymous

yes

27. 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|

28. anonymous

okay...

29. 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

30. Michele_Laino

what do you get?

31. Michele_Laino

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

32. anonymous

-22/256 =22/-32

33. Michele_Laino

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

34. anonymous

thank you

35. Michele_Laino

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

36. anonymous

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

37. anonymous

676/-64

38. Michele_Laino

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

39. Michele_Laino

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

40. anonymous

thank you

41. Michele_Laino

please wait, we have to solve part C

42. anonymous

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

43. 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|

44. anonymous

yes

45. Michele_Laino

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

46. anonymous

i understand so far

47. Michele_Laino

:)

48. Michele_Laino

that's all!

49. anonymous

thank you!

50. Michele_Laino

:)

51. anonymous

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

52. anonymous

@Michele_Laino

53. 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.