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Finding the Maximum or Minimum of a Quadratic  Problem 1
Carl Horowitz
Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upperlevel math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
So one of the applications of a quadratic equation is to find a maximum or minimum of a relationship and one of the most common relationships we're looking at is something being thrown up and then coming back down and looking for the maximum height and when that maximum height occurs.
So what we're looking at for this particular example is a rocket that's being shot up, so it's height in meters t seconds after being launched is given by this equation and the question we're trying to answer is when will it reach it's maximum height?
So by looking at this quadratic equation, I know that the graph is going to be a parabola facing downwards because my coefficient is negative, so when our coefficient is negative, it tells us this graph is facing downwards which means at some point we're going to have a maximum spot. That maximum occurs at the vertex, so in order to answer this question we have to find the vertex.
We have two different ways of doing that. We can complete the square or we can just use b over 2a. b over 2a tends to be easier so let's just go with that. So b over 2a is just going to be 80 over 2 times 8, or basically 80 over 16 because our negative cancel. 80 over 16 is just going to give us 5.
So what we have found is the x coordinate of our vertex, however in this problem, our x is our t so what we really found is the t coordinate of the vertex, t is time so what we have found is the time when reach our maximum, when we reached the vertex. So that's what the question is asking for throw in the unit and we have 5 seconds as the times we reach our maximum height.
The other portion of this question is saying what will its maximum height be? We've already found one coordinate of the vertex, we found the time portion, what we want to find is the height portion, so all we have to do is take this 5 and plug it into this equation to figure out what our maximum height will be.
It's easy enough so we end up taking 8 times 25 plus 80 times 5 plus 25 and we end up with 225 feet. So these are the applications of the vertex in a word problem and just a quick reminder of other things we could do with type of problem, we could say when is the rocket going to be 100 feet? You would set it equal to 100 everything over and solve it out.
So basically a quadratic equation representing something we can solve for a number of different things. We can solve for when it hits the ground, its maximum height, when it reach the maximum height all sorts of different things. The vertex is only going to give us extreme, only going to give us either the maximum or the the minimum if our parabola is facing the other direction.
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Carl Horowitz
B.S. in Mathematics University of Michigan
He knows how to make difficult math concepts easy for everyone to understand. He speaks at a steady pace and his stepbystep explanations are easy to follow.
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Sample Problems (4)
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Finding the Maximum or Minimum of a Quadratic
Problem 1 6,187 viewsA rocket is shot up so its height in meters 't' sec. after launch is given by h(t) = 8t² + 80t + 25.
a) When will it reach its max height?b) What is its max height? 
Finding the Maximum or Minimum of a Quadratic
Problem 2 3,055 viewsA town is fencing in a playground whihc is bordered on one side by a building. The plan is to use 640 ft of fencing on the other three sides.
What should the dimension of the playground be if the enclosed area is to be maximized? 
Finding the Maximum or Minimum of a Quadratic
Problem 3 218 views 
Finding the Maximum or Minimum of a Quadratic
Problem 4 219 views
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