Leam with an example
1 answer:
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<h3>Answer:</h3>
y = 8x² -4x +4
<h3>Explanation:</h3>I find it quick and easy to let a graphing calculator do quadratic regression on the given points. It gives the coefficients of the equation directly. (See attached.)
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If you want to do this "by hand," you can substitute each of the x and y value pairs into the equation to get three linear equations in a, b, and c. These are generally easy to solve, as the "c" variable can be eliminated right away.
<em>For x = 2</em>
... 28 = a·2² +b·2 +c = 4a +2b +c
<em>For x = 4</em>
... 116 = a·4² +4b +c = 16a +4b +c
<em>For x = -3</em>
... 88 = a·(-3)² +b·(-3) +c = 9a -3b +c
Subtracting the first and third equations from the second, we have ...
... (16a +4b +c) -(4a +2b +c) = (116) -(28) ⇒ 12a +2b = 88
... (16a +4b +c) -(9a -3b +c) = (116) -(88) ⇒ 7a +7b = 28
Dividing the first of these reduced equations by 2, and the second by 7, we have ...
- 6a +b = 44
- a +b = 4
Subtracting the second of these from the first gives ...
... (6a +b) -(a +b) = (44) -(4) ⇒ 5a = 40
From which we find
... a = 8
... b = 4 -a = 4 -8 = -4
We choose the first of the original equations to find c:
... c = 28 -4a -2b = 28 -4·8 -2(-4)
... = 28 -32 +8 = 4
With (a, b, c) = (8, -4, 4), the equation of the parabola is ...
... y = 8x² -4x +4