Question

Create a system of equations and use algebra to write a quadratic equation for each set of three points that lie on a parabola. (-1,2),(4,27)(-3,20)

Answer 1

y = 2x² -x -1

Explanation:

You want the equation for the quadratic curve through the points (-1, 2), (4, 27), and (-3, 20).

Equations

The standard form equation for the parabola is ...

y = ax² +bx +c

Substituting for x and y, we have ...

• 2 = a(-1)² +b(-1) +c = a - b + c
• 27 = a(4)² +b(4) +c = 16a +4b +c
• 20 = a(-3)² +b(-3) +c = 9a -3b +c

Solution

Subtracting the first equation from the other two gives ...

• 25 = 15a +5b
• 18 = 8a -2b

These can be reduced to standard form:

• 5 = 3a +b
• 9 = 4a -b

Adding these two equations, we have ...

14 = 7a ⇒ a = 2

Substituting into the first gives ...

5 = 3(2) +b

5 -6 = b = -1

And substituting into the very first equation, we have ...

2 = 2 -(-1) +c

2 -3 = c = -1

Quadratic

The quadratic that has these points on its graph is ...

y = 2x² -x -1

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Additional comment

We like to use a calculator to solve systems of linear equations using the matrix row reduction technique. This is shown in the second attachment. The same calculator can also do quadratic regression, writing the equation directly from the three point coordinates.

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