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

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Answer 1 · 2016-11-06T20:50:17

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

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:

Here is a plot of 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 $y = x^2 + 2x - 2$