Graph a Quadratic in Vertex From Easy Med Hard

so what if you need to graph a quadric equation in vertex form what is an example of a easy medium and hard well this video is exactly for you because that is exactly what we are going to do so let's go and take a look at this first one the easy one and the reason why it's easy because it's already in vertex form okay now you recognize there's no number in front of the parentheses so therefore it's going to be a one but that's going to be nice because there's no vertically stretching or compressing or flipping about the XX AIS right all we need to simply be considerate about is going to be our vertex what exactly is our vertex and a lot of times students will make the simple mistake of saying it's 1 -2 but just remember the formula here is xus H so a better way that we could write this equation is y = a x minus a -1 2us 2 right because minus a negative 1 is the same thing as positive you would agree but now it matches up nicely with this formula so therefore we can identify here the ver vertex to be a -1 -2 and then to graph It All We simply going to do is plot that point we know since a is positive the orientation or the graph is going to open up and we're not worried about any stretching or compression if we wanted to find some points we could plug it in but we'll get to that in the next problem and usually the main points that I always like to plug in you know usually like having x equals 0 to find the uh y intercepts or plug zero or Y in for zero to find the X intercepts you know you can play around with that pick points from there now let's go and take a look at this one because sometimes students are like all right that's pretty easy right it's already in vertex form I like when it's in vertex form what about when it's not in vertex form well therefore you'd have to complete the square but I'll do that in the hard one for a hard completing the square problem because this problem wouldn't be very difficult for completing the square once you've learned it what about a problem like this like we don't have a middle term to even complete the square on so how could I graph this using this model if it has a middle term you're want to complete the square if it doesn't have a middle term you just need to understand how could this actually Rewritten in that format and this is what we can do right because technically when we look at this to this graph the x coordinate of the vertex is going to be zero because would you agree x - 0 is just X and therefore it' be squared so therefore the vertex here is going to be a0 comma 4 but here's where we have the problem because our a is no longer a positive 1 it's nice when it's a positive 1 now we have a -2/3 crap what the heck are we going to do with thega -2/3 we know the negative is the graph is going to open down right we got that and but the 2/3 since it's actually less than one that is actually going to be a horizontal stretch right or I'm sorry it's a going to be a vertical compression it's on the outside so it's a vertical compression but to what like stretch like if we're just sketching it a lot of times that'd be fine but like how can I identify like another Point here to work with well the one thing that's important about this is typically we always like to have multiple points but when you're looking at a parabola you have this axis of symmetry that goes through the vertex if you pick one point to the left of the axis symmetry you can utilize that exact same point to the right of the axis of symmetry so what I need to do is Select at least one value that'd be rather easy for me to be able to evaluate here because we have a fraction right so what I want to do is I want to pick a value of x that when I Square it it's going to be divisible by three that will eliminate my fraction right because if I have something in my denominator evenly divides into I no longer need the denominator so what I recognize here is if I chose one three does not divide into one or one squar right which is one if I did two two squar is four three does not divide a four if I do three 3^ squ is n so why don't we go ahead and pick a point here and I'll do what about when Y is equal to 3 okay so now I have x equal to 3 right 3^2 is 9 3 divides the 9 3 * 3 * -2 is going to be a -6 -6 + 4 is going to be a -6 + 4 is going to be a -2 so now I have a point here which is going to be a 3 comma -2 so let's go ahead and graph this with the remaining work that I have we have a Vertex at 04 so 1 2 3 4 and then I have a point at 3 -2 1 2 3 down two well here's the cool thing I plugged in three and I got a -2 what would happen if I plugged in A3 right well a-3 squ would still give me nine right so I also have that point as well that's where that axis symmetry because the axis symmetry in this case is going to be the Y AIS and voila I now have a graph that I'm just sketching right it might not be perfect from in that case and actually you know what since I talked about doing that let's go and find another Point here but I did find three points let's find one more point I think the easiest point to find here is going to be when X is equal to Zer so let's go and figure out what that point would be so when x equals z that's going to be the Y intercept here so that's going to be one 1^ s is 1 one things so that's a negative one so actually the graph kind of looks like that okay so we did find a point that's going to be one so we have zero comma sorry negative 1 all right let's go and find at least one more Point rather than the vertex right now this is a the hard problem and the reason why that's hard because it's in standard form and yes you can go ahead and graph this using standard form you can identify the axis of symmetry using opposite of B divide by 2A and then you could plug that in to find the vertex and then you're kind of at the same point here but I think it's important to when you're understanding how to graph using vertex form to make sure that you can convert this using complete in the Square because graphing quadratics is not the only time we use complete in the Square it is a very very important process to be able to make sure you know how to do so let's go ahead and get started when I have an A in front front of my or I have anything in front of my X squ I'm going to need to factor that out okay so since a three is not divisible by two I had to leave it as a fraction and again check my work always check your work if you multiply the two back to both of these I'm still going to get a 2x^2 minus 3 right now what I need to do is find the value that completes the square here that is going to be your B and remember to find your B you're just going to take your sorry to find your C what you're going to do is you're going to take your B / by two and then Square it so C which is going to be the value that completes the square that makes this a perfect square trinomial is going to be B / 2^ 2ar so in my case I have A3 / by two and I need to square that all right well be careful because dividing by two is the same thing as multiplying by 1/2 right so technically what this is is a -34 squared which is going to be a 9 916 all right so what I'm going to do is I'm going to add that value inside the parentheses but I'm also going to subtract it outside right because you whatever you do on one side you just you have to do it on the other side or you can add and subtract on the same side so I'm actually going to draw this in a different color so therefore you can see what I did here so that is going to be a 9 over a 16 all right so now I'm going to have y = a 2 * x^2 - 3 X X and that's going to be plus a 9 over a 16 and then I'm going to minus a 1 right because those there and then minus a 9 over 16 because I added here so I have to subtract it and then here's the important thing remember this is actually not a 9 over 16 this is a 9 over 16 that's been multiplied by two so that means I need to multiply this 9 over 16 by two as well all right so now we need to identify what the perfect square trinomial is and guess what guys whatever your B / 2 is that's going to be your perfect square trinomial so y = a 2 * a xus uh a 34s co^ squar let's go and do this work here so that's going to be a 1 over8 I can rewrite A1 is 8 over 8 so therefore it' be 8 over 8 minus a 9 over 8 so therefore that's going to be a Nega 17 over 8 so therefore my vertex is going to be a positive 34 minus a 17 over 8 comma -7 over8 and I do not want to go ahead and find another Point that's not going to be very cool but I can go ahead and estimate right so let's see we have 34 so let's say that's one so 34 would be here and then a -7 over 8 think about it this way guys that's going to be like 8 * 2 is 16 if you had eight so that'd be two and just a little bit more than two and then this graph is going to be a stretch of two so therefore I'll leave it from on there but if I had more work you could definitely go ahead and plug in another value that would work to be able to figure out another point to be honest with you you could also pick out a when X is equal to zero if you plug zero in for X that or I'm sorry a 34s into X well that's going to give you your x axis from on there so yeah we don't need that that' be A7 over 8 but what number would I pick if I wanted to I don't know we're going to stop that video from here and there you go that's what the graph is going to look like at least a good sketch