Hello and welcome. In this mathematics post, I cover the topic of determining quadratic functions given a table of values.

Math text is done with rendered Latex with QuickLatex.com.

## What Is A Quadratic Function?

A quadratic function is a type of mathematical function of the form:

where `x`

is the independent variable, `f(x)`

is the dependent variable that depends on the value of `x`

. The coefficients `a`

, `b`

and `c`

are numeric values.

A linear function such as `g(x) = 3x`

has `g(x)`

being increased by 3 every time `x`

is increased by 1. The increase is consistent and does not change every time `x`

is increased by 1. With a quadratic function, every time `x`

is increased by 1, the amount that `f(x)`

increases by does not stay the same.

**Examples Of Quadratic Functions**

Let's take a look at some quadratic functions along with their graphs (Screenshots from Desmos). Different quadratic functions will their own shapes in how they open up and down.

## Using Table Of Values To Determine Quadratic Functions

There are times when all you have is just have a table of values. You need to determine if a quadratic function does fit the points from the table of values. If so, what is the quadratic function that does fit the points?

**Example One**

Here is a table of values for the quadratic function in this example.

x | y |
---|---|

-2 | -1 |

-1 | -2 |

0 | -1 |

1 | 2 |

2 | 7 |

3 | 14 |

We need to take a look at the first differences to see if there are the same. First differences takes a y-value minus the previous y-value if it exists.

x | y | First Differences |
---|---|---|

-2 | -1 | N/A |

-1 | -2 | -1 |

0 | -1 | +1 |

1 | 2 | +3 |

2 | 7 | +5 |

3 | 14 | +7 |

As the first differences are not the same throughout, the function that fits through these points is **not** a linear function. Let's take a look at the second differences.

x | y | First Differences | Second Differences |
---|---|---|---|

-2 | -1 | N/A | N/A |

-1 | -2 | -1 | N/A |

0 | -1 | +1 | +2 |

1 | 2 | +3 | +2 |

2 | 7 | +5 | +2 |

3 | 14 | +7 | +2 |

The second differences are the same which means that the function that fits these points is a quadratic function of the form . What are the values for the coefficients though?

For the coefficient value of `a`

, it is half of the second difference. In this example the value of `a`

is half of 2 which is 1.

The value of `c`

is the y-intercept. This is the value of `y`

when the value of `x`

is 0. For this example, the `c`

value is -1.

Determining the value of `b`

requires algebra along with substituting values from a point from the table in the quadratic function general form. I will use the point (1, 2).

The quadratic equation that fits the points in the table of values is .

**Example Two - Not A Quadratic Function**

Not every table of values is associated with a quadratic function as shown in this example.

x | f(x) |
---|---|

-2 | -6 |

-1 | -3 |

0 | -2 |

1 | -3 |

2 | -7 |

Taking the first differences and the second differences gives this modified table.

x | f(x) | First Differences | Second Differences |
---|---|---|---|

-2 | -6 | N/A | N/A |

-1 | -3 | +3 | N/A |

0 | -2 | +1 | -2 |

1 | -3 | -1 | -2 |

2 | -7 | -4 | -3 |

The second differences are not the same. If you find one instance where the second difference is not the same in a table of values, then a quadratic function does not fit through all the points. There is no such quadratic function here.

**Example Three**

Notice how the last two examples include a y-intercept where there is a y-value when x is equal to 0. This example will show a case of a table of values where there is no y-intercept given.

x | g(x) |
---|---|

1 | 0 |

2 | -5 |

3 | -14 |

4 | -27 |

5 | -44 |

As `x`

increases the amount for `g(x)`

decreases. Let's take a look at the first differences and the second differences.

x | g(x) | First Differences | Second Differences |
---|---|---|---|

1 | 0 | N/A | N/A |

2 | -5 | -5 | N/A |

3 | -14 | -9 | -4 |

4 | -27 | -13 | -4 |

5 | -44 | -17 | -4 |

The second differences is a constant negative 4. The value for `a`

in is half of -4 which is -2. We now need to determine the values for `b`

and `c`

for the quadratic function that can fit through these five points. This has to be done with a system of equations.

Select two `(x, y)`

points from the table and set up to equations. The first equation uses the x and y value pair from (1, 0) and the second equation uses the x and y pair from the point (2, -5).

**First Equation**

**Second Equation**

To solve this system of equations, you can use the elimination method, a substitution method, or even use matrices from linear algebra. For this example, I will use the substitution method as we can isolate the variable `b`

(or `c`

) in the first equation.

Start from the first equation:

Use this value of `b`

into the `b`

in the second equation. Afterwards, solve for c.

In the second equation:

The value for `c`

is found which can be used to find the value for `b`

. I am using equation one to find `b`

.

A lot has happened in this example. Here is a summary.

x | g(x) | First Differences | Second Differences |
---|---|---|---|

1 | 0 | N/A | N/A |

2 | -5 | -5 | N/A |

3 | -14 | -9 | -4 |

4 | -27 | -13 | -4 |

5 | -44 | -17 | -4 |

As the second difference is -4, this makes the value for `a`

in the quadratic function half of -4 which is negative 2.

After some algebra, the quadratic function that fits through the (x, y) points above is .

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