Question

How do you write a quadratic function in standard form whose graph passes through points (1,2), (-2,5), (3,-20)?

Answer 1

Use the standard form:

`y = ax^2+bx+c" [1]"`

and the 3 given points to create rows in an augmented matrix.
Solve the matrix for `a,b, and c`
Substitute the values into equation [1]

Explanation:

Use the point `(1,2)`in equation [1]:

`a(1^2) + b(1) + c = 2`

`a + b + c = 2`

This gives us the first row:

#[ (1,1,1,|,2) ]#

Use the point `(-2,5)` in equation [1]:

`a(-2^2) + b(-2) + c = 5`

`4a - 2b + c = 5`

This gives us the second row:

#[ (1,1,1,|,2), (4,-2,1,|,5) ]#

Use the point `(3,-20)` in equation [1]:

`a(3^2) + b(3) + c = -20`

`9a+ 3b + c = -20`

This gives us the third row:

#[ (1,1,1,|,2), (4,-2,1,|,5), (9,3,1,|,-20) ]#

Perform elementary two operations until we obtain an identity matrix on the left.

`R_2-4R_1 to R_2`:

#[ (1,1,1,|,2), (0,-6,-3,|,-3), (9,3,1,|,-20) ]#

`R_3-9R_1toR_3`:

#[ (1,1,1,|,2), (0,-6,-3,|,-3), (0,-6,-8,|,-38) ]#

`R_3-R_2toR_3`:

#[ (1,1,1,|,2), (0,-6,-3,|,-3), (0,0,-5,|,-35) ]#

`-1/5R_3toR_3`:

#[ (1,1,1,|,2), (0,-6,-3,|,-3), (0,0,1,|,7) ]#

`R_2+3R_3toR_2`:

#[ (1,1,1,|,2), (0,-6,0,|,18), (0,0,1,|,7) ]#

`-1/6R_2toR_2`:

#[ (1,1,1,|,2), (0,1,0,|,-3), (0,0,1,|,7) ]#

`R_1-R_3toR_1`

#[ (1,1,0,|,-5), (0,1,0,|,-3), (0,0,1,|,7) ]#

`R_1-R_2toR_1`

#[ (1,0,0,|,-2), (0,1,0,|,-3), (0,0,1,|,7) ]#

We have an identity matrix on the left, therefore the solution is in the column vector on the right:

`a = -2, b = -3 and c = 7`

Substitute these values into equation [1]:

`y = -2x^2-3x+7" [2]"`

To prove that this is the correct solution, here is a graph of the 3 points and equation [2]: