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anonymous

  • one year ago

when graphing a quadratic equation -x^2+2x=5 what would be the points I would plot for it. I know the axis of symmetry is 1.

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  1. jtvatsim
    • one year ago
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    It's a good practice to plot at least 3 points for a quadratic equation. It is always really helpful to plot the point for the axis of symmetry (plug in x =1 into the equation to get the y value). The other two points are up to you to pick. I might pick x = 0 since that is easy, and maybe x = 2. But it is really a matter of personal taste. There isn't a right or wrong way to pick these two points. :)

  2. jtvatsim
    • one year ago
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    But, wait, do you have an equal sign in your question?

  3. jtvatsim
    • one year ago
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    I may have misread that. :)

  4. anonymous
    • one year ago
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    yes but when I do this problem i am stumped i'm told all of my work was wrong look for number 5 that is the problem i am having

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  5. jtvatsim
    • one year ago
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    Number 5 looks fine to me, do you mean number 6?

  6. anonymous
    • one year ago
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    sorry my mistake yes it is number six not number 5

  7. jtvatsim
    • one year ago
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    OK, let me see...

  8. jtvatsim
    • one year ago
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    OK, first thing, it looks like you added 5 to both sides. You needed to subtract 5 from both sides.

  9. anonymous
    • one year ago
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    okay but where do I subtract the five on one of the sides

  10. jtvatsim
    • one year ago
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    So, you have -x^2 + 2x = 5. We will set this equal to 0 (that is usually the best way to do it). So, -x^2 + 2x = 5 -5 -5 -x^2 + 2x - 5 = 0 Does that make sense?

  11. anonymous
    • one year ago
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    yes it does then what would be the next step

  12. jtvatsim
    • one year ago
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    OK, probably we should find the axis of symmetry, which you did and found it was x = 1.

  13. jtvatsim
    • one year ago
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    That's correct. So let's find the y-value that goes with that by plugging in x = 1.

  14. jtvatsim
    • one year ago
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    -(1)^2 + 2(1) - 5 = -1+2-5=-4 So the vertex is (1,-4). Good so far?

  15. anonymous
    • one year ago
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    yes. Now what is next?

  16. jtvatsim
    • one year ago
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    Well, we want to see if there are any solutions, that make this equation 0, but there probably won't because of the -x^2. You can see the shape of the graph by plugging in two random points. I would suggest x = 0 and x = 2 since they are small and close to the axis of symmetry.

  17. anonymous
    • one year ago
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    I need to have two point of the right and left pf the vertex and the solution is 1

  18. jtvatsim
    • one year ago
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    OK, you've almost got it. Yes, it's a good idea to have two points of the right and left of the vertex. However, the "solution" is not 1. There are none for this graph. Let me explain using the picture.

  19. jtvatsim
    • one year ago
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    |dw:1437167305706:dw|

  20. jtvatsim
    • one year ago
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    That -4 should be by the vertex, but you get the point. :)

  21. jtvatsim
    • one year ago
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    So, what we have done is we have drawn the graph of -x^2 +2x - 5 by plotting the points.

  22. jtvatsim
    • one year ago
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    Now, in order to see if there are ANY solutions to -x^2 + 2x - 5 = 0, we want to see if the graph ever crosses the x-axis. Wherever the graph crosses the x-axis is a solution.

  23. jtvatsim
    • one year ago
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    But, from our picture we see that the graph never crosses the x-axis. So there are no solutions.

  24. jtvatsim
    • one year ago
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    It's a tricky idea, so do you have any questions about that?

  25. jtvatsim
    • one year ago
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    Step 1) Set equation = 0. Step 2) Draw the graph of the quadratic by finding axis of symmetry and choosing points. Step 3) See if the graph crosses the x-axis. Wherever it does this is a solution. If it doesn't there are no solutions.

  26. anonymous
    • one year ago
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    where would I plot the points that's what is confusing me the most.

  27. jtvatsim
    • one year ago
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    OK, that's a good question. For the axis of symmetry, we found x = 1, and y = -4. So we plot this.

  28. jtvatsim
    • one year ago
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    |dw:1437167688817:dw|

  29. jtvatsim
    • one year ago
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    That dot is the first point plotted for the axis of symmetry.

  30. jtvatsim
    • one year ago
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    Does that make sense so far? There's two more for us to plot. :)

  31. anonymous
    • one year ago
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    I get that part easy it just the other points that i don't get

  32. jtvatsim
    • one year ago
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    OK, that's fine. It's actually really easy once you get the hang of it.

  33. jtvatsim
    • one year ago
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    So, I'm going to pick a random number. But I'll pick an easy one. How about x = 0? Then, I get -0^2 +2*0 - 5 = 0 + 0 - 5 = -5. So, I have x = 0, y = -5. as another point.

  34. jtvatsim
    • one year ago
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    |dw:1437167834263:dw|

  35. anonymous
    • one year ago
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    that will be the same for x=2. correct?

  36. jtvatsim
    • one year ago
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    Yes, because of the axis of symmetry! So, x = 2, y = -5.

  37. jtvatsim
    • one year ago
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    |dw:1437167900952:dw|

  38. jtvatsim
    • one year ago
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    We can plot more points if we want, but I'm pretty convinced that this is what the graph looks like...

  39. jtvatsim
    • one year ago
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    |dw:1437167946786:dw|

  40. jtvatsim
    • one year ago
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    Of course, you'll want to be a better artist than me... :)

  41. anonymous
    • one year ago
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    I agree with you my lines are a lot worse than you have seen.

  42. jtvatsim
    • one year ago
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    haha! :D

  43. anonymous
    • one year ago
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    Thank you so very much and sorry about my grammar with my sentence.

  44. jtvatsim
    • one year ago
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    Ah no worries, as long as I can help I don't mind. :) Nice work! You were very close to the right answer.

  45. anonymous
    • one year ago
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    Thanks I have a hard time understanding math since I have autism which makes it harder for me to understand how to do certain things.

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