We ‘re asked to graph the following equation yttrium equals 5x squared subtraction 20x plus 15. so let me get my fiddling scratch diggings out. So it ‘s y is equal to 5x squared minus 20x plus 15. now there ‘s many ways to graph this. You can equitable take three values for ten and number out what the corresponding values for yttrium are and equitable graph those three points. And three points actually will determine a parabola. But I want to do something a little bit more interesting. I want to find the places. so if we imagine our axes. This is my x-axis. That ‘s my y-axis. And this is our curvature. So the parabola might look something like this. I want to first name out where does this parabola intersect the x-axis. And as we have already seen, intersecting the x-axis is the same thing as saying when it does this when does y adequate 0 for this trouble ? Or another way of saying it, when does this 5x squared minus 20x plus 15, when does this equal 0 ? so I want to figure out those points. And then I besides want to figure out the point precisely in between, which is the vertex. And if I can graph those three points then I should be all set with graphing this parabola. So as I just said, we ‘re going to try to solve the equality 5x squared subtraction 20x plus 15 is equal to 0. now the beginning thing I like to do whenever I see a coefficient out here on the adam squared term that ‘s not a 1, is to see if I can divide everything by that term to try to simplify this a little spot. And possibly this will get us into a factor-able phase. And it does look like every term hera is divisible by 5. So I will divide by 5. so I ‘ll divide both sides of this equation by 5. And so that will give me — these cancel out and I ‘m left with ten squared minus 20 over 5 is 4x. Plus 15 over 5 is 3 is equal to 0 over 5 is just 0. And now we can attempt to factor this left-hand side. We say are there two numbers whose product is incontrovertible 3 ? The fact that their intersection is plus tells you they both must be positive. And whose sum is damaging 4, which tells you well they both must be veto. If we ‘re getting a negative sum here. And the one that credibly jumps out of your thinker — and you might want to review the video on factoring quadratics if this is not therefore fresh — is a negative 3 and negative 1 seem to work. minus 3 times negative 1. negative 3 times negative 1 is 3. negative 3 plus negative 1 is minus 4. so this will factor out as adam minus 3 times adam minus 1. And on the right-hand side, we hush have that being equal to 0. And immediately we can think about what x ‘s will make this construction 0, and if they make this formulation 0, well they ‘re going to make this expression 0. Which is going to make this formula adequate to 0. And then this will be true if either one of these is 0. so adam minus 3 is adequate to 0. Or ten subtraction 1 is equal to 0. This is genuine, and you can add 3 to both sides of this. This is true when ten is equal to 3. This is true when ten is peer to 1. So we were able to figure out these two points veracious over here. This is adam is equal to 1. This is ten is adequate to 3. then this is the steer 1 comma 0. This is the target 3 comma 0. And indeed the stopping point one I want to figure out, is this point correct over here, the vertex. now the vertex always sits precisely big h dab between the roots, when you do have roots. sometimes you might not intersect the x-axis. So we already know what its x-coordinate is going to be. It ‘s going to be 2. And nowadays we good have to substitute back in to figure out its y-coordinate. When adam equals 2, yttrium is going to be equal to 5 times 2 squared subtraction 20 times 2 plus 15, which is equal to — let ‘s see, this is equal to 2 squared is 4. This is 20 minus 40 plus 15. indeed this is going to be negative 20 plus 15, which is equal to negative 5. so this is the point 2 comma negative 5. And so now we can go back to the exert and actually plot these three points. 1 comma 0, 2 comma veto 5, 3 comma 0. So let ‘s do that. So first I ‘ll do the vertex at 2 comma damaging 5, which is right there. And now we besides know one of the times it intersects the x-axis is at 1 comma 0. And the other meter is at 3 comma 0. And now we can check our answer. And we got it right.