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.

Question

jtvatsim · Answer

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 · Answer

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

jtvatsim · Answer

I may have misread that. :)

anonymous · Answer

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

jtvatsim · Answer

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

anonymous · Answer

sorry my mistake yes it is number six not number 5

jtvatsim · Answer

OK, let me see...

jtvatsim · Answer

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

anonymous · Answer

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

jtvatsim · Answer

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 · Answer

yes it does then what would be the next step

jtvatsim · Answer

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

jtvatsim · Answer

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

jtvatsim · Answer

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

anonymous · Answer

yes. Now what is next?

jtvatsim · Answer

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 · Answer

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

jtvatsim · Answer

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 · Answer

|dw:1437167305706:dw|

jtvatsim · Answer

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