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# Common Core Standards: Math

#### The Standards

# High School: Functions

### Linear, Quadratic, and Exponential Models HSF-LE.B.5

**5. Interpret the parameters in a linear or exponential function in terms of a context.**

Your students have been solving linear and exponential functions for what feels like centuries, and it's about to come to an end. As one final test of everything they've learned, you ask them a simple question. "Okay, class. We have *y* dollars in revenue for every *x* packets of gum sold and *y* = 0.95*x*. If *x* equals 20, how much will *y* equal?"

Jimmy, a brilliant young student of yours, beams up at you and answers, "It'll equal 19 packets of gum." Your heart sinks to the floor, and you realize your mistake all along.

It's not enough to understand *how* to solve linear and exponential functions. Sure, it's useful, but it just won't cut it. Students should know what equations actually mean when applied to certain contexts, not just how to solve for *x*.

Understanding an equation's context is important not only so that students know in what units to report their answers (in dollars and not packets of gum), but also in order to take in data and make use of it. This means that given a particular context, students should be able to understand trends, make predictions, and extrapolate from the mathematical functions they're given.

Some students make these connections quickly and effortlessly, and others might find it a little more difficult to do so. One possible way to assist those struggling students is to assign a clear meaning to each variable so that they know that *y* always means dollars and *x* always means packets of gum.

In more complex problems such as exponentials and polynomials, it may be useful to break down the problem so that it's clearly understood *what* is changing by *how much* for every *what*. Translating the equation into words or vice versa may help understand the equation in terms of the overall context. (For instance, every additional packet of gum sold, denoted by *x*, increases the revenue *y* by 0.95 dollars. That's what the equation *y* = 0.95*x* ultimately means.)