How To Use Formulas: An Example

Consider the formula y=a(x-h)^2+k.  Knowing nothing about it except that it is an amalgamation of symbols and a number, we can still extract some information if we need to.  For instance we can attempt to solve for any of the variables.  So, solving for a, yields a = \frac{(y-k)}{(x-h)^2}.  Or if we know the value of four out of five of the variables, we can find the value of the fifth.  So if y=0,x=0,h=2, and k=-1, we first substitute the four values to get 0=a(0-2)^2-1 and then solve for a to get a= \frac{1}{4}.  Note we could have used the equation for a that we derived at the beginning of this paragraph.  We did all of this without any hint of where the formula came from or what it is about.  Rule of thumb: If you know all but one of the variables in a formula, you can find the other one most of the time.

If we know where the formula comes from and the meaning of its variables,we see it in a different light. y=a(x-h)^2+k is the “standard” form for a second degree polynomial function.  Its graph has the shape of a parabola.  Now x stands for  the independent variable or the input to the function or the first coordinate of a point on the graph and y stands for the dependent variable or the output of the function or the second coordinate of a point on the graph. a,h,k are parameters.  This means that if we specify values for them, we have specified the location and behavior of the graph and the minimum or maximum value of its output.  h,k are the coordinates of the vertex (the highest or lowest point) and a specifies behavior.  If a<0, the graph is concave down (sheds water) and has a maximum value and if a>0 the graph is concave up (holds water) and has a minimum value.  If |a|>1 the graph is skinny (rises more rapidly) and If |a|<1 the graph is wide (rises more slowly).  Now the formula “talks” to us.  It is telling a story. y=\frac{1}{4}(x-2)^2-1 is saying “I’m wide and hold water and have a vertex at (2,-1).

Here is a sample problem.  Find the equation of a second degree polynomial with vertex  at (2,-1) and which passes through the origin.  Using the standard form for the parabola, y=a(x-h)^2+k and understanding what its variable mean we already have y=a(x-2)^2-1.  All that remains is to find a.  But the phrase, “passes through the origin” is just a sneaky way of saying that the point x=0,y=0 is a point on the graph that satisfies the formula.  Now we know four out of the five variables, y=0,x=0,h=2, and k=-1, so we can solve for a as in the first paragraph.  The desired equation is y=\frac{1}{4}(x-2)^2-1  and we know that it is wide and has a minimum value of -1.

Conclusion:  Sometimes it is sufficient to treat a formula like any other algebraic expression, but most of the time it is useful even necessary to understand all of its parts.

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About jrh794

I am a sixty-five year old math instructor at Southern Oregon University. I taught at the College of the Siskiyous in Weed California for twenty-six years. Prior to that I worked as a computer programmer, carpenter and in various other jobs. I graduated from Rice University in 1967 and have a MS in Operations Research from Stanford. In the past I have hand-built a stone house and taken long solo bicycle tours. Now I ride my mountain bike and play golf for recreation.
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