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

Nov 6, 2016

$y = {x}^{2} + 2 x - 2$

#### Explanation:

There are two standard forms for a parabola:

Type 1 opens up or down:

$y = a {x}^{2} + b x + c$

Type 2 opens left or right:

$x = a {y}^{2} + b y + c$

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:

$1 = a {\left(1\right)}^{2} + b \left(1\right) + c \text{ [1]}$
$- 3 = a {\left(- 1\right)}^{2} + b \left(- 1\right) + c \text{ [2]}$
$1 = a {\left(- 3\right)}^{2} + b \left(- 3\right) + c \text{ [3]}$

Move the coefficients in front of the variables:

$1 = a + b + c \text{ [1]}$
$- 3 = a - b + c \text{ [2]}$
$1 = 9 a - 3 b + c \text{ [3]}$

Subtract equation [1] from equation [2]

$1 = a + b + c \text{ [1]}$
$- 4 = - 2 b \text{ [2]}$
$1 = 9 a - 3 b + c \text{ [3]}$

$1 = a + b + c \text{ [1]}$
$2 = b \text{ [2]}$
$1 = 9 a - 3 b + c \text{ [3]}$

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

$1 = a + 2 + c \text{ [1]}$
$1 = 9 a - 3 \left(2\right) + c \text{ [3]}$

$- 1 = a + c \text{ [1]}$
$7 = 9 a + c \text{ [3]}$

Subtract equation [1] from equation [3]:

$- 1 = a + c \text{ [1]}$
$8 = 8 a \text{ [3]}$

$- 1 = a + c \text{ [1]}$
$1 = a \text{ [3]}$

Substitute 1 for a in equation [1]:

$c = - 2$

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

$1 = {\left(1\right)}^{2} + 2 \left(1\right) - 2$
$- 3 = {\left(- 1\right)}^{2} + 2 \left(- 2\right) - 2$
$1 = {\left(- 3\right)}^{2} + 2 \left(- 3\right) - 2$

$1 = 1$
$- 3 = - 3$
$1 = 1$

This checks.

The equation is $y = {x}^{2} + 2 x - 2$