anonymous
  • anonymous
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.
Algebra
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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jtvatsim
  • jtvatsim
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. :)
jtvatsim
  • jtvatsim
But, wait, do you have an equal sign in your question?
jtvatsim
  • jtvatsim
I may have misread that. :)

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anonymous
  • anonymous
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
1 Attachment
jtvatsim
  • jtvatsim
Number 5 looks fine to me, do you mean number 6?
anonymous
  • anonymous
sorry my mistake yes it is number six not number 5
jtvatsim
  • jtvatsim
OK, let me see...
jtvatsim
  • jtvatsim
OK, first thing, it looks like you added 5 to both sides. You needed to subtract 5 from both sides.
anonymous
  • anonymous
okay but where do I subtract the five on one of the sides
jtvatsim
  • jtvatsim
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?
anonymous
  • anonymous
yes it does then what would be the next step
jtvatsim
  • jtvatsim
OK, probably we should find the axis of symmetry, which you did and found it was x = 1.
jtvatsim
  • jtvatsim
That's correct. So let's find the y-value that goes with that by plugging in x = 1.
jtvatsim
  • jtvatsim
-(1)^2 + 2(1) - 5 = -1+2-5=-4 So the vertex is (1,-4). Good so far?
anonymous
  • anonymous
yes. Now what is next?
jtvatsim
  • jtvatsim
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.
anonymous
  • anonymous
I need to have two point of the right and left pf the vertex and the solution is 1
jtvatsim
  • jtvatsim
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.
jtvatsim
  • jtvatsim
|dw:1437167305706:dw|
jtvatsim
  • jtvatsim
That -4 should be by the vertex, but you get the point. :)
jtvatsim
  • jtvatsim
So, what we have done is we have drawn the graph of -x^2 +2x - 5 by plotting the points.
jtvatsim
  • jtvatsim
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.
jtvatsim
  • jtvatsim
But, from our picture we see that the graph never crosses the x-axis. So there are no solutions.
jtvatsim
  • jtvatsim
It's a tricky idea, so do you have any questions about that?
jtvatsim
  • jtvatsim
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.
anonymous
  • anonymous
where would I plot the points that's what is confusing me the most.
jtvatsim
  • jtvatsim
OK, that's a good question. For the axis of symmetry, we found x = 1, and y = -4. So we plot this.
jtvatsim
  • jtvatsim
|dw:1437167688817:dw|
jtvatsim
  • jtvatsim
That dot is the first point plotted for the axis of symmetry.
jtvatsim
  • jtvatsim
Does that make sense so far? There's two more for us to plot. :)
anonymous
  • anonymous
I get that part easy it just the other points that i don't get
jtvatsim
  • jtvatsim
OK, that's fine. It's actually really easy once you get the hang of it.
jtvatsim
  • jtvatsim
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.
jtvatsim
  • jtvatsim
|dw:1437167834263:dw|
anonymous
  • anonymous
that will be the same for x=2. correct?
jtvatsim
  • jtvatsim
Yes, because of the axis of symmetry! So, x = 2, y = -5.
jtvatsim
  • jtvatsim
|dw:1437167900952:dw|
jtvatsim
  • jtvatsim
We can plot more points if we want, but I'm pretty convinced that this is what the graph looks like...
jtvatsim
  • jtvatsim
|dw:1437167946786:dw|
jtvatsim
  • jtvatsim
Of course, you'll want to be a better artist than me... :)
anonymous
  • anonymous
I agree with you my lines are a lot worse than you have seen.
jtvatsim
  • jtvatsim
haha! :D
anonymous
  • anonymous
Thank you so very much and sorry about my grammar with my sentence.
jtvatsim
  • jtvatsim
Ah no worries, as long as I can help I don't mind. :) Nice work! You were very close to the right answer.
anonymous
  • anonymous
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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