For this question were given a set of data points, and they are the path of a ball thrown where X is the horizontal distance. And why is the height we're asked to find a quadratic equation in the form Y equals a X squared plus BX plus C that contains these points. To do that, we can create a system of equations and sulfur are unknowns, which are a BNC. Yeah, And then I'll use matrices to solve the system of equations. So to create our system of equations, I'm going to use what we know. So the first thing we know is that when x zero y is five. So if I plug in zero for X into our quadratic equation, we get zero plus zero plus C equals five. So our equation is C equals five. The second data point is when X is 15. Why is 9.6 so plugging in 15 for X, we get to 25 a plus 15 b first see is 9.6 and our third data point is when X is 30 we get 900 a plus. 30 b plus C is 12.4. Now we have our system of equations, and we can put it into an augmented matrix to solve for a BNC. All right, I'm going to start with this equation. I want this one to be last, so I'm going to do 225 a plus 15 b plus one C is 9.6. My second equation will be 900. A plus, 30 b plus one C is 12.4, and my third equation will be zero a plus zero B plus one C is five now. I can use elementary row operations. Start eliminating some terms. I want this to be zero. So to do that, I'm going to multiply the first row by negative four. That will give me 225 in the first cell. So when I add it to Row two, okay, zero. Yeah, so my first row stays the same. My second row becomes to 25 times negative four, which is negative. 900 plus 900 0 negative. Four times 15 plus 30 is negative. 30 and negative four times one plus one is negative. Three negative four times 9.6 plus 12.4 is negative. 26. My third row will stay the same now to get this into row echelon form. I want this to be zero. I'm sorry, I want this to be one and I want this to be one. So I'm going to divide Row one by 2 25 and I'm going to divide row to buy negative 30 that will give US one 15. Divided by 2 25 is 250.66 Repeating one divided by 2 25 is 250.44 REPEATING 9.6 divided by 2 25 is 0.4266 Repeating My second row will have zero negative 30 divided by negative 30 is one negative three divided by negative 30 0.1 and negative 26. Divided by negative. 30 is 300.0 0.866 Repeating. My third row will stay the same. 0015 Now that I have my matrix and row echelon form, I can use back substitution to solve for A B and C. So my first column contains a coefficients. My second is the B, and my third is the sea. So this third row here already tells us that C equals five. My second row tells us that B plus 0.1 c equals 0.866 repeating. And if I plug in that C is five we can solve for B. So we get that. B is approximately 0.37. It's 0.366 repeating, and this first equation tells us that a plus 0.66 b plus 0.44 c equals 0.4266 And if I plug in what I know, which is that B is 0.37 C is five. I can solve for a and I get a equals negative 0.4 So those are the coefficients to our quadratic formula, and we get y equals negative 0.4 x squared plus 0.37 x plus five. So we can graph that, and it looks like a parabola. Now. The third part of our problem says to graphically approximate the maximum height of the ball and the point at which the ball hits the ground. The maximum height. The ball is approximately here. Remember that are why access is the height and the point at which the ball hits the ground. Is that the horizontal distance when the height is zero? So it's approximately here. We could approximate those at about 14 ft, maybe in 105 ft. To solve this analytically. The maximum height is going to be at the Vertex, which is when X equals negative B over two. A. We can plug in what we know, which is V as 0.37 and A is negative 0.4 and we get that. The maximum height is when the horizontal distance is 46.25 ft. So that's about here 46.25 But we want to know the maximum height, and so that's why. So what's why, when X is 46.25 we just plug 46.25 into X here, and we get that Y is approximately 13.56 beach. Now we can find the point at which the ball hits the ground. That's going to be the X intercept, and we can find that by setting the equation equal to zero, and finally we can use the quadratic equation to solve this for X. So X is going to be negative B plus or minus the square root a B squared, minus four ac over to a. And when we solve that for X, we'll get that X is approximately one. Oh, four point. I believe it is 0.47 So the horizontal distance would be one Oh, four point 5 ft when the ball hits the ground and those do match up with what we thought they were graphically.