Given three points (-1,4) (1,6) (2,10) how do you write a quadratic function in standard form with the points?

1 Answer
May 2, 2017

Solve a system of three equations in three unknowns.

Explanation:

Let us assume that the axis of symmetry for the parabola is vertical.

The standard equation of a parabola with a vertical axis is #y = ax^2 + bx + c#, where a, b, and c are real numbers -- with #a ne 0#.

Using the coordinates of the above points, we know that...
#a(-1)^2 + b(-1) + c = 4#
This simplifies to #a - b + c = 4#.

We also have:
#a1^2 + b(1) + c = 6#
That is, #a + b + c = 6#.

And finally:
#a(2)^2 + b(2) + c = 10#
#4a + 2b + c = 10#

The three equations give us the following system, which has a unique solution (a, b, c):

#a - b + c = 4#
#a + b + c = 6#
#4a + 2b + c = 10#.

While we may solve these using any method, it will be convenient to subtract equation (1) from equation (2).

#a + b + c = 6#
#a - b + c = 4# (subtract)

Change the signs and add:
#a + b + c = 6#
#-a + b - c = -4# (add)

Adding gives us:
#2b = 2#
#b = 1#.

Our system is now reduced to just two unknowns. Putting the known value of b into the three equations gives:

#a - 1 + c = 4#
#a + 1 + c = 6#
#4a + 2 + c = 10#.

Simplify to:
#a + c = 5#
#4a + c = 8#.

Subtract the first from the second to obtain the equation:
#3a = 3#
#a = 1#

Substitute a = 1 into the equation #a + c = 5# and obtain #c = 4#.

The equation of our parabola has a = 1, b = 1, and c = 4.

Therefore, it is
#y = x^2 + x + 4#
Check, and you will find that all three points fit the equation.