so today we want to look at some ways to graph a quadratic function without using a full blown table of values and hopefully before watching this video you've had a chance to do a bit of exploration with quadratic functions where we begin with a basic parabola y equals x squared as you see it here and then what we're going to do is we're going to change some of the parameters or numbers in various parts of the equation so over here the first thing I'm going to do is I'm going to change the value of a so instead of having 1 times x squared what would happen if I had 2 times x squared or 3 times x squared etc etc so as I do that I'll notice that if I change it to 2 the graph is getting skinnier and the higher a gets the skinnier it gets let's just go back towards an a value of 1 where it's back to the regular one if I go less than that the graph gets wider all of the Y values are being multiplied by the value of a causing them to go above or below where they were previously which is then causing the graph to stretch or compress if I go to a negative value for AOL notice that the graph is opening downwards it's a reflection across the x axis as you can think of it and again the same kind of wideness narrowness issue happens as i change the value of a so what we're seeing by exploring with our parabola is multiplying by a does two things first of all if a is positive the parabola opens up if a is negative the parabola opens down and then if the value of a isn't larger than one and in fact if it's negative the same thing happens if we think of the positive about negative so I'm gonna say absolute value which just means the positive of whatever the a value is if that number is greater than one we get a vertical stretch the graph stretches vertically by a times if the value of a is less than one but bigger than zero it can't be zero otherwise it reduces to a linear function we get a vertical compression by a times okay so multiplying the function by a constant a changes whether it opens up or down and it changes the wideness or skinniness of the graph all right the next parameter we're going to influence is a number added or subtracted 2x on inside of the brackets which I'm calling P in this animation they're calling it B so let's take B and move it to be a more positive value positive one let's go up to B equals two now you'll notice what's happened is the graph has shifted two units to the right and my equation is X minus 2 you might be wondering we're saying B is to remember the minus is there so if you just slot take the B away and slop the two and you get X minus 2 so we see X minus 2 squared we're thinking of the B value just being that number if we go backwards to a negative B value negative 3 all right so we're slotting negative 3 into here X minus negative 3 makes it X plus 3 as you see it here and what is that done it shifted the graph 3 units to the left so as you play around with B you can see what it's doing is shifting the graph translating the graph left and right it's keeping it the same shape it's keeping a congruent to what the original one was but it's shifting it left and right all right so P if greater than zero translates the graph P units to the right and just to make this clear because this is a little bit confusing sometimes for people if I write X minus two squared what I'm saying is that T is two which is a shift two to the right if P is negative then I'm translating the graph P units left and again just to make that clear if I have y equals x plus one squared because my standard way of writing this is set up is X minus I have to think of this as X minus negative one squared and therefore P is negative one which is to the left the final influence we want to look at is adding a constant Q to the end of the equation and in my applet that's represented by C so let's go to a positive values of C and what you'll see is the graph is translating vertically up and down if I go up to a C value of three x squared plus three my graph has shifted upwards three units if I go to a negative value negative one let's say if the constant is negative one at the end my graph vertically translates one unit downwards so if Q is a positive number graph translates up Q units if Q is negative the graph translate down Q units so we can summarize these transformations into a standard form equation of the quadratic function if you have y equals x squared multiplying x squared by a will do a vertical stretch or compression as well as open the graph up and down if you add or subtract into the X you will do a horizontal translation left or right and if you add a constant on at the end Q that will vertically translate the graph up or down furthermore since the vertex of x squared was 0 0 then the vertex of the standard form if it's been shifted T units horizontally and Q units vertically would become P comma Q a is the vertical stretch compression direction of opening and P and Q are the horizontal and vertical translations so if I were to look at my animation and take my a value and change it to my original graph x squared vertically stretches because a is positive by a factor of two which means each of the y-values see the original Y value here of one it goes up twice as far to to the Y value here of four goes up twice as far to eight its vertically expanded stretched think of it stretched vertically this way by a factor of two if I change my B to negative three X minus negative three shows up as X plus three I've shifted my graph three units to the left and if I change C let's say to negative one I'm vertically translating my graph one unit down alright now what you're gonna need to do is take what we just talked about and be able to graph a parabola fairly accurately on paper so I'm gonna graph this parabola the equation that I have here with three points of accuracy and indicate the asked for properties so I'm just gonna write my equation down below so it's a bit bigger my equation you'll notice is given in standard form which means in the form of what we just talked about a number multiplied to the X part X's has something added or subtract to it it on the inside and there's a bit of constant Q on the outside and we know what a each of AP and Q does okay so a is 2 which means the graph is going to open up because we have a positive value of a it's gonna be narrower than a normal problem because a is more than one p is negative one because we think of it as X minus negative one which means we're gonna translate graph one to the left and Q is negative 3 which means we're translating the graph 3 down and what the translation information gives us is the vertex we're not vertex at 0 0 we are one unit left and 3 units down right there so my vertex is negative 1 negative 3 which as we indicated before is the values of the P and Q parameter the axis of symmetry runs through the vertex vertically so the axis of symmetry is this reflection line so we know if that's x equals negative 1 now we want points of accuracy we know the graph opens up we know it's narrower than normal so what I like to do is just take a value of x what are two units over from the vertex so from negative 1 I'm going to go over to x equals 0 and I'm going to plug x equals 0 into the equation alright so I'm just going to talk this out x equals 0 so 0 plus 1 is 1 squared is 1 2 times 1 is 2 2 minus 3 is negative 1 so I have a point of accuracy there which conveniently turns out to be the y-intercept which I was asked for so that's negative 1 and because of symmetry I know that if this point is on the graph then there's a point on the other side of the axis of symmetry like so and then I can complete my sketch the minimum value of the graph it's a minimum because the graph opens up the minimum value is the y-value there's a minimum value of negative 3 the domain of a parabola of a quadratic function is all reals and the range as we can see is a low Y value of negative 3 and anything above that so Y is greater than or equal to negative 3 or Y is a real number all right now that we have some familiarity with standard form let's work in Reverse we have information about the ground and we want to come up with its standard form equation so find the parabola in standard form that has a vertex of negative 3/5 so I'm going to do a little sketch of this so I can understand it vertex here it has an x intercept of negative 1 so the graph passes through this point now because of symmetry we know that it would have to have another x intercept an equal distance on the other side there so my graph looks something like so so it opens downwards alright so we're going to build the equation it's really what we're being asked to do what is the equation that represents that parabola so the first thing that to put in is the vertex because the vertex is the values of P and Q y equals a X minus negative 3 which becomes plus 3 is we know plus 5 now we know that a is gonna turn out to be a negative because the graph is opening down but the issue is there's lots of graphs opening downwards from this vertex lots of graphs right it's wide ones skinny ones narrower ones but we want the one that passes through these x-intercepts all right so what we're told is that if the x intercept is negative 1 the graph passes through negative 1 comma 0 and we know that if a graph passes through a point that point must work in the equation of that relation so if this point is on the graph it has to work in this equation for x and y so let's plug in Y is 0 x is negative 1 and then we got to figure out what he would have to be to make that work so this is two squared that's for a plus five solving for a negative for a is five so therefore a is negative five over four and so we can complete the equation by putting that into that position so our standard form equation is y equals negative five over four X plus three squared plus five so I've typed this into my calculator a X minus P squared plus Q and my graph should look like the sketch I had previously vertex of negative 3/5 and it has an x intercept of negative one so things are looking good