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.

- anonymous

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- 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

But, wait, do you have an equal sign in your question?

- jtvatsim

I may have misread that. :)

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- 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

Number 5 looks fine to me, do you mean number 6?

- anonymous

sorry my mistake yes it is number six not number 5

- jtvatsim

OK, let me see...

- jtvatsim

OK, first thing, it looks like you added 5 to both sides. You needed to subtract 5 from both sides.

- anonymous

okay but where do I subtract the five on one of the sides

- 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

yes it does then what would be the next step

- jtvatsim

OK, probably we should find the axis of symmetry, which you did and found it was x = 1.

- jtvatsim

That's correct. So let's find the y-value that goes with that by plugging in x = 1.

- jtvatsim

-(1)^2 + 2(1) - 5 = -1+2-5=-4 So the vertex is (1,-4). Good so far?

- anonymous

yes. Now what is next?

- 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

I need to have two point of the right and left pf the vertex and the solution is 1

- 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

|dw:1437167305706:dw|

- jtvatsim

That -4 should be by the vertex, but you get the point. :)

- jtvatsim

So, what we have done is we have drawn the graph of -x^2 +2x - 5 by plotting the points.

- 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

But, from our picture we see that the graph never crosses the x-axis. So there are no solutions.

- jtvatsim

It's a tricky idea, so do you have any questions about that?

- 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

where would I plot the points that's what is confusing me the most.

- jtvatsim

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

- jtvatsim

|dw:1437167688817:dw|

- jtvatsim

That dot is the first point plotted for the axis of symmetry.

- jtvatsim

Does that make sense so far? There's two more for us to plot. :)

- anonymous

I get that part easy it just the other points that i don't get

- jtvatsim

OK, that's fine. It's actually really easy once you get the hang of it.

- 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

|dw:1437167834263:dw|

- anonymous

that will be the same for x=2. correct?

- jtvatsim

Yes, because of the axis of symmetry!
So, x = 2, y = -5.

- jtvatsim

|dw:1437167900952:dw|

- jtvatsim

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

- jtvatsim

|dw:1437167946786:dw|

- jtvatsim

Of course, you'll want to be a better artist than me... :)

- anonymous

I agree with you my lines are a lot worse than you have seen.

- jtvatsim

haha! :D

- anonymous

Thank you so very much and sorry about my grammar with my sentence.

- 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

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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