### 5.1 Just do it!

Lesson materials located below the video overview.

Let’s get straight into it!

EXAMPLE 102: Find a quadratic function that fits the following data:

That is, find a quadratic function $$y=ax^{2}+bx+c$$ that passes through the three points $$\left({2},{7}\right)$$, $$\left({5},{10}\right)$$, and $$\left({7},{3}\right)$$.

Answer: The best thing to do is to just write down the answer! Here it is:

$$y=7 \cdot \frac {(x-5)(x-7)}{(-3)(-5)} + 10 \cdot \frac{(x-2)(x-7)}{3\cdot(-2)} + 3 \cdot \frac{(x-2)(x-5)}{5\cdot 2}$$

If one were to expand this out we’d see that this is indeed a quadratic function. But, of course, this is not the issue in one’s mind right now. Perhaps the question “From where does this formula come and what is it doing?” is more pressing!

To understand this meaty formula start by plugging in the value $$x=2$$. (Do it!) Notice that the second and third terms are designed to vanish at $$x=2$$ and so we have only to contend with the first term: $$7\cdot \frac{(x-5)(x-7)}{(-3)(-5)}$$. When $$x=2$$ the numerator and the denominator match (the denominator was designed to do this) so that this term becomes:

$$7 \cdot 1$$

which is the value $$7$$  we want from the table we were given.

For the value $$x=5$$ only the middle term $$10 \cdot \frac{(x-2)(x-7)}{3 \cdot (-2)}$$ “survives” and has value$$10 \cdot \frac {3\cdot (-2)}{3 \cdot (-2)}=10$$ for $$x=5$$.

In the same way, for the value $$x=7$$ only the third term is non-vanishing and has value: $$3\cdot \frac{5 \cdot 2}{5 \cdot 2} = 3$$.

Thus the quadratic $$y=7 \cdot \frac {(x-5)(x-7)}{(-3)(-5)} + 10 \cdot \frac{(x-2)(x-7)}{3\cdot(-2)} + 3 \cdot \frac{(x-2)(x-5)}{5\cdot 2}$$ does indeed produce the values $$7$$, $$10$$ and $$3$$ for the inputs $$2$$, $$5$$ and $$7$$, respectively!

COMMENT: If one so desired we can expand this to write: $$y=-\frac{9}{10}x^{2}+\frac{219}{30}x-4$$.      □

Despite the visual complication of the formula one can see that its construction is relatively straightforward:

1. Write a series of numerators that vanish in turn at all but one of the desired inputs.
2. Create denominators that cancel the numerators when a specific input is entered.
3. Use the desired $$y$$-values as coefficients.

COMMENT: Do you see the connection here to “finding interesting $$x$$-values”. Each term is focusing on making two of the three $$x$$-values give the same output of zero.

As another example, here’s a quadratic that passes though the points $$A=\left({3},{87}\right)$$, $$B=\left( {10},{\pi}\right)$$ and $$\left({35},{\sqrt{2}}\right)$$:

$$y=87\frac{(x-10)(x-35)}{(-7)(-32)}+\pi\frac{(x-3)(x-35)}{(7)(-28)}+\sqrt{2}\frac{(x-3)(x-10)}{32\cdot25}$$

CHECK: Put in $$x=3$$. Do you get the output $$87$$? Also, put in $$x=10$$ and then $$x=35$$.

I choose not to expand this, but a tiny bit of simplification would be fine:

$$y=\frac{87}{224}(x-10)(x-35) -\frac{\pi}{196}(x-3)(x-35) + \frac{\sqrt{2}}{800}(x-3)(x-10)$$.

This looks slightly less scary.

EXAMPLE 103: Find a quadratic that fits the data:

Answer: $$y=a\frac{(x-2)(x-3)}{(-1)(-2)} + b\frac{(x-1)(x-3)}{1\cdot (-1)}+c\frac{(x-1)(x-2)}{2 \cdot 1}$$ □

Recall that all solutions to the practice problems appear in the COMPANION GUIDE to this QUADRATICS course.

PRACTICE 104: Find a quadratic that fits the data:

### IF YOU NEED TO SIMPLIFY ALL THE WAY …  (but first ask yourself it is really necessary!)

It can be done and it actually isn’t as bad as you first think it might be.

For example, consider the data:

A quadratic that fits this data is:

$$y=3\frac{(x+1)(x-3)}{(3)(-1)} + 6\frac{(x-2)(x-3)}{(-3)(-4)} + 10\frac{(x-2)(x+1)}{(1)(4)}$$.

A tiny bit of simplifying gives:

$$y=-(x+1)(x-3)+\frac{1}{2}(x-2)(x-3)+\frac{5}{2}(x-2)(x+1)$$.

To handle the fractions, it might be easiest to put the terms over a common denominator:

$$y=\frac{-2(x+1)(x-3) +(x-2)(x-3)+5(x-2)(x+1)}{2}$$

Expanding each product gives:

$$y=\frac{-2(x^{2}-2x-3)+(x^{2}-5x+6)+5(x^{2} -x-2)}{2}$$.

We now see:

The $$x^{2}$$ terms are: $$\frac{-2+1+5}{2}x^{2}=2x^{2}$$.

The $$x$$ terms are: $$\frac{4-5-5}{2}x=-3x$$.

The constant term is: $$\frac{6+6-10}{2}=1$$.

Thus the quadratic that fits the data is: $$y=2x^{2}-3x+1$$.

PRACTICE 105: Find a quadratic that goes through the points $$\left({-3},{-14}\right)$$, $$\left({2},{1}\right)$$ and $$\left( {3},{-2}\right)$$. For fun, simplify your answer all the way!

PRACTICE 106: Find a quadratic that fits the data:

PRACTICE 107: Something interesting happens if one tries to find a quadratic that fits the points $$\left({2},{7}\right)$$, $$\left({3},{9}\right)$$ and $$\left( {6},{15}\right)$$.

b) What happened and why?

PRACTICE 108: Something goes wrong if one tries to find a quadratic that fits the data $$\left({1},{0}\right)$$, $$\left({1},{-2}\right)$$ and $$\left( {-1},{-1}\right)$$.

a) Try to write a quadratic that fits this data.

b) What goes wrong and why?

OPTIONAL CHALLENGE: Find an equation of the form $$x=ay^{2}+by+c$$ that fits this data!

PRACTICE 109: Why stop at quadratics?

Here is some data and here is a formula that fits it:

$$y=4\frac{(x-6)(x-8)(x-9)}{(-3)(-5)(-6)}+8\frac{(x-3)(x-8)(x-9)}{(3)(-2)(-3)}-3\frac{(x-3)(x-6)(x-9)}{(5)(2)(-1)}+12\frac{(x-3)(x-6)(x-8)}{(6)(3)(1)}$$.

a) Write down a formula that fits the data:

b) Here is a table of data that spells my name. (Do you understand the connection?)

I wrote a formula that fits the data perfectly and then had a computer algebra system do the simplifying for me. Here’s the matching formula that spells my name:

$$p(x)=\frac{83}{24}x^{4}-\frac{331}{12}x^{3}+\frac{1657}{24}x^{2}-\frac{647}{12}x+10$$.

[So putting in $$x=0$$ gives $$p(0)=10$$, and putting in $$x=1$$ gives $$p(1)=1$$, and so on.]

What is the formula that spells your name?

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