Question

How do you write a quadratic function in standard form whose graph passes through points (3,15), (-4,121/2), (1/3, -7/3)?

Answer 1

Because you specified a "function", we must use the standard form:
`y = ax^2+bx+c`
Use the 3 points to write 3 equations and then use linear algebra techniques to solve for `a, b, and c`

Explanation:

NOTE: If you did not specify a "function" then there would be two equations; the second one would be the standard form:

`x = ay^2+bx+c`

Returning to the function.

Using the point `(3,15)`:

`a(3)^2 + b(3)+c = 15`

This makes the first line of the augmented matrix:

`[(9,3,1,|,15)]`

Using the point `(-4,121/2)`

`a(-4)^2 + b(-4)+c = 121/2`

`16a - 4b+c = 121/2`

Multiply by 2:

`32a - 8b+2c = 121`

Add this line to the augmented matrix:

#[(9,3,1,|,15), (32,-8,2,|,121)]#

Use the point `(1/3,-7/3)`

`a(1/3)^2 + b(1/3)+c = -7/3`

`1/9a + 1/3b +c = -7/3`

Multiply by 9:

`a + 3b +9c = -21`

Add this line to the top of the augmented matrix:

#[(1,3,9,|,-21), (9,3,1,|,15), (32,-8,2,|,121)]#

Perform elementary row operations:

`R_2-9R_1toR_2`

#[(1,3,9,|,-21), (0,-24,-80,|,204), (32,-8,2,|,121)]#

`R_3-32R_1toR_3`

#[(1,3,9,|,-21), (0,-24,-80,|,204), (0,-104,-286,|,793)]#

`R_2/-4`

#[(1,3,9,|,-21), (0,6,20,|,-51), (0,-104,-286,|,793)]#

`3R_3+52R_2toR_3`

#[(1,3,9,|,-21), (0,6,20,|,-51), (0,0,182,|,-273)]#

`R_3/182`

#[(1,3,9,|,-21), (0,6,20,|,-51), (0,0,1,|,-3/2)]#

`R_2-20R_3toR_2`

#[(1,3,9,|,-21), (0,6,0,|,-21), (0,0,1,|,-3/2)]#

`R_2/6`

#[(1,3,9,|,-21), (0,1,0,|,-7/2), (0,0,1,|,-3/2)]#

`R_1-9R_3toR_1`

#[(1,3,0,|,-15/2), (0,1,0,|,-7/2), (0,0,1,|,-3/2)]#

`R_1-3R_2toR_1`

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

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

`y = 3x^2-7/2x-3/2" "larr` The answer

Check:

`3(3)^2-7/2(3)-3/2=15`
`3(-4)^2-7/2(-4)-3/2=121/2`
`3(1/3)^2-7/2(1/3)-3/2=-7/3`

`15=15`
`121/2=121/2`
`-7/3=-7/3`

This checks.