How do you write a quadratic function in standard form whose graph passes through points (1,2), (-2,5), (3,-20)?

1 Answer
Jun 12, 2017

Use the standard form:

#y = ax^2+bx+c" [1]"#

and the 3 given points to create rows in an augmented matrix.
Solve the matrix for #a,b, and c#
Substitute the values into equation [1]

Explanation:

Use the point #(1,2)#in equation [1]:

#a(1^2) + b(1) + c = 2#

#a + b + c = 2#

This gives us the first row:

#[ (1,1,1,|,2) ]#

Use the point #(-2,5)# in equation [1]:

#a(-2^2) + b(-2) + c = 5#

#4a - 2b + c = 5#

This gives us the second row:

#[ (1,1,1,|,2), (4,-2,1,|,5) ]#

Use the point #(3,-20)# in equation [1]:

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

#9a+ 3b + c = -20#

This gives us the third row:

#[ (1,1,1,|,2), (4,-2,1,|,5), (9,3,1,|,-20) ]#

Perform elementary two operations until we obtain an identity matrix on the left.

#R_2-4R_1 to R_2#:

#[ (1,1,1,|,2), (0,-6,-3,|,-3), (9,3,1,|,-20) ]#

#R_3-9R_1toR_3#:

#[ (1,1,1,|,2), (0,-6,-3,|,-3), (0,-6,-8,|,-38) ]#

#R_3-R_2toR_3#:

#[ (1,1,1,|,2), (0,-6,-3,|,-3), (0,0,-5,|,-35) ]#

#-1/5R_3toR_3#:

#[ (1,1,1,|,2), (0,-6,-3,|,-3), (0,0,1,|,7) ]#

#R_2+3R_3toR_2#:

#[ (1,1,1,|,2), (0,-6,0,|,18), (0,0,1,|,7) ]#

#-1/6R_2toR_2#:

#[ (1,1,1,|,2), (0,1,0,|,-3), (0,0,1,|,7) ]#

#R_1-R_3toR_1#

#[ (1,1,0,|,-5), (0,1,0,|,-3), (0,0,1,|,7) ]#

#R_1-R_2toR_1#

#[ (1,0,0,|,-2), (0,1,0,|,-3), (0,0,1,|,7) ]#

We have an identity matrix on the left, therefore the solution is in the column vector on the right:

#a = -2, b = -3 and c = 7#

Substitute these values into equation [1]:

#y = -2x^2-3x+7" [2]"#

To prove that this is the correct solution, here is a graph of the 3 points and equation [2]:

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