Question

How do you find an equation in standard form of the parabola passing through the points (1,1), (-1, -3), (-3, 1)?

Answer 1

Please see the explanation for steps leading to the answer:
`y = x^2 + 2x - 2`

Explanation:

There are two standard forms for a parabola:

Type 1 opens up or down:

`y = ax^2 + bx + c`

Type 2 opens left or right:

`x = ay^2 + by + c`

Plot the 3 points:

The points look like they fit the first type.

Write 3 equations by substituting the 3 points into the type 1 form:

`1 = a(1)^2 + b(1) + c" [1]"`
`-3 = a(-1)^2 + b(-1) + c" [2]"`
`1 = a(-3)^2 + b(-3) + c" [3]"`

Move the coefficients in front of the variables:

`1 = a + b + c" [1]"`
`-3 = a - b + c" [2]"`
`1 = 9a - 3b + c" [3]"`

Subtract equation [1] from equation [2]

`1 = a + b + c" [1]"`
`-4 =-2b" [2]"`
`1 = 9a - 3b + c" [3]"`

`1 = a + b + c" [1]"`
`2 =b" [2]"`
`1 = 9a - 3b + c" [3]"`

Substitute 2 for b into equations [1] and [3]:

`1 = a + 2 + c" [1]"`
`1 = 9a - 3(2) + c" [3]"`

`-1 = a + c" [1]"`
`7 = 9a + c" [3]"`

Subtract equation [1] from equation [3]:

`-1 = a + c" [1]"`
`8 = 8a" [3]"`

`-1 = a + c" [1]"`
`1 = a" [3]"`

Substitute 1 for a in equation [1]:

`c = -2`

Confirm that the equation: `y = x^2 + 2x - 2` fits all 3 points:

`1 = (1)^2 + 2(1) - 2`
`-3 = (-1)^2 + 2(-2) - 2`
`1 = (-3)^2 + 2(-3) - 2`

`1 = 1`
`-3 = -3`
`1 = 1`

This checks.

The equation is `y = x^2 + 2x - 2`