Question

find an equation in the form y=ax^2+bx+c for the parabola passing through the points. (2,28),(4,116),(-3,88)

Answer 1

Answer:

y = 8x² -4x +4

Explanation:

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.

For x = 2

... 28 = a·2² +b·2 +c = 4a +2b +c

For x = 4

... 116 = a·4² +4b +c = 16a +4b +c

For x = -3

... 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