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