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

1 Answer
Nov 6, 2016

Answer:

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:

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#