# The Vertex and Axis of Symmetry - Problem 3

###### Transcript

So here is another example of how we can use Algebra to solve problems that we find in our everyday life. Let’s take a look. A soccer ball’s height in feet follows the path described by h(x) is equal to 48x minus 16x² where x is the time in seconds. After how long does the soccer ball reach its maximum height? What is that height?

So let’s go ahead and let’s underline the relevant information and see exactly what we want to find. Well we know that we have a function height of (x)is equal to 48x minus 16 times x². We also know that what we want to find is we want to find the time in seconds where the ball reaches it’s maximum height. We also want to find out what is that height?

So like we should do in every word problem when we’re solving it, let’s go ahead and draw a diagram. So here we have our soccer player kicking the soccer ball and we know that it’s going to follow a trajectory. It’s going to go up, and it’s going to go back down before it hits the ground. What we want to find is we want to find this maximum height, and we wanted to find the number of seconds that it takes to get there. So this makes sense, we’re looking at a parabola and what we want to find is we want to find the vertex.

We know that given this coefficient is negative, we know that our parabola opens downwards which make sense. Remember we always need to think logically how these problems work. So what we want to find is we want to find our vertex. Now remember to find our x value we set up a simple equation. We know that x is equal to the opposite of b all over 2a, so our b value is this one right here, 48 so we know that the opposite of 48 over 2a, our a value is right here, so -2 times 16 is going to give us our x value of our vertex.

So if we simplify this, what we get is we get the opposite of 48 over the opposite of 32, remember the negatives cancel out, we’re left with 3/2. This makes sense.

So we know that our time that the soccer ball reaches the maximum height is 3/2 of seconds, so 1 ½ seconds. What we can do is we can plug it in right here and then using this information, we can find our y value which is going to give us this maximum height.

So plug in this value back into our function what we get is that 48 times 3/2 minus 16 times 3/2 squared, remember it’s really important to look back at our function to figure out exactly what we’re plugging in and making sure that we’ve got it all right. So 48 times 3/2 is equal to 72 and what we’re going to do is we’re going to first square this 3/2, so we have -16 times 3/2 squared is 9/4 and we can simplify this. No need to multiply everything out.

We know that 4 goes into 16 four times, 4 times 9 is 36, it looks a little messy, but you get the point 72 minus 36 and we have our maximum height of 36.

So remember when you’re solving a word problem especially with quadratics, it’s really important to draw out a diagram to think and to plug the numbers back in and make sure that you’re coming up with the exact answers that it’s asking.