Unit 8, Lesson 24
Using Quadratic Equations to Model Situations and Solve Problems
Algebra 1
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A golf ball is shot straight up into the air so that its height above the ground, in meters, is given by , where represents the number of seconds after the ball is launched.
A camera is on a device that was on the ground 6 seconds before the ball was launched, and it rises at a constant rate so that it is 60 meters above the ground when the ball is hit.
Certain real-world situations can be modeled by quadratic functions, and these functions can be represented by equations. Sometimes, all the skills we have developed are needed to make sense of these situations. When we have a mathematical model and the skills to use the model to answer questions, we are able to gain useful or interesting insights about the situation.
Suppose we have a model for the height of a launched object, , as a function of time since it was launched, , defined by . We can answer questions such as these about the object’s flight:
(An expression in standard form can help us with this question. Or, we can evaluate to find the answer.)
(When an object hits the ground, its height is 0, so we can find the zeros using one of the methods we learned: graphing, rewriting the equation in factored form, completing the square, or using the quadratic formula.)
(We can rewrite the expression in vertex form, or we can use the zeros or a graph of the function to find the vertex.)
Sometimes, relationships between quantities can be effectively communicated with graphs and expressions rather than with words. For example, these graphs represent a linear function, , and a quadratic function, , with the same variables for their inputs and outputs.
If we know the expressions that define these functions, we can use our knowledge of quadratic equations to answer questions such as:
(Yes. We can see that their graphs intersect at a couple of places.)
(To find out, we can write and solve this equation: . The solution provides the -values for the intersection points, and the -values can be found by substituting the solutions for in either original function.)