Question

Find an equation whose graph is a horizontally opening parabola that passes through (20, -3), (12,1), and (33,-2)?

Answer 1

The equation is:

`x = -5y^2-12y+29`

Here is a graph of the parabola and the 3 points:

Explanation:

The standard form for a horizontally opening parabola is:

`x = ay^2+by+c" [1]"`

Substituting the point `(12,1)` into equation [1]:

`a(1)^2+b(1) + c = 12`

`a+b+c = 12`

This equation translates into the first row of an augmented matrix as:

#[ (1,1,1,|,12) ]#

Substituting the point `(20, -3)` into equation [1]:

`a(-3)^2+b(-3) + c = 20`

`9a-3b+c=20`

This equation translates into the second row of the augmented matrix as:

#[ (1,1,1,|,12), (9,-3,1,|,20) ]#

Substituting the point `(33,-2)` into equation [1]:

`a(-2)^2+b(-2) + c = 33`

`4a-2b+c=33`

This equation translates into the second row of the augmented matrix as:

#[ (1,1,1,|,12), (9,-3,1,|,20), (4,-2,1,|,33) ]#

Perform elementary row operations, until an identity matrix is obtained on the left:

`R_2-9R_1toR_2`

#[ (1,1,1,|,12), (0,-12,-8,|,-88), (4,-2,1,|,33) ]#

`R_3-4R_1toR_3`

#[ (1,1,1,|,12), (0,-12,-8,|,-88), (0,-6,-3,|,-15) ]#

`R_3-1/2R_2toR_3`

#[ (1,1,1,|,12), (0,-12,-8,|,-88), (0,0,1,|,29) ]#

`R_2+8R_3toR_2`

#[ (1,1,1,|,12), (0,-12,0,|,144), (0,0,1,|,29) ]#

`-1/12R_2toR_2`

#[ (1,1,1,|,12), (0,1,0,|,-12), (0,0,1,|,29) ]#

`R_1-R_2toR_1`

#[ (1,0,1,|,24), (0,1,0,|,-12), (0,0,1,|,29) ]#

`R_1-R_3toR_1`

#[ (1,0,0,|,-5), (0,1,0,|,-12), (0,0,1,|,29) ]#

We have an identity matrix on the left, therefore, we can read the solutions on the right, `a = -5, b = -12, and c = 29`

The equation is:

`x = -5y^2-12y+29`