Question

write the equation of a parabola in standard form that contains the points (0, 0), (-1, -5), and (2, -2)

Answer 1

Answer:

  y = -2x² + 3x

Step-by-step explanation:

The y-intercept of 0 tells you the constant is zero.

You can find the coefficients "a" and "b" in the form

  y = ax² +bx +0

by substituting the data points that are not the y-intercept.

For x=-1, ...

  -5 = a(-1)² +b(-1) = a - b

For x=2, ...

  -2 = a(2)² +b(2) = 4a +2b

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These two equations can be solved by your favorite method. Here's one way:

You can divide the second equation by 2 and add the first:

  (-2)/2 +(-5) = (4a +2b)/2 +(a -b)

  -6 = 3a

  -2 = a

Then you can use either equation to find b.

  -5 = -2 -b . . . substitute for "a" in the first equation

  b = 3 . . . . . . add b+5

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Now, you know the quadratic is

  y = -2x² +3x

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Alternate solution

You can let your graphing calculator or spreadsheet program tell you the equation of a quadratic regression using these points. The attached shows one such result. (It is the same as above.)

Answer 2

Answer: y = -2x² + 3x

Step-by-step explanation:

The standard form of a quadratic equation is: y = ax² + bx + c

Input the x, y coordinates provide to create three equations. Then solve the system of equations

(0, 0) → 0 = a(0)² + b(0) + c

            0 =   0    +   0   + c

            0 = c

(-1, -5) and c = 0 → -5 = a(-1)² + b(-1) + 0

                             -5 =    a     +   -b

(2, -2) and c = 0 → -2 = a(2)² + b(2) + 0

                             -2 =  4a    +  2b

-5 =   a -   b  →  2(-5 = a  -  b)  →  -10 = 2a - 2b

-2 = 4a + 2b → 1(-2 = 4a + 2b) → -2 = 4a + 2b

                                                   -12 = 6a

                                                   ÷6   ÷6  

                                                    -2 = a

-5 = a - b

-5 = -2 - b

-3 = -b

3 = b

a = -2, b = 3, c = 0  →   y = -2x² + 3x + 0