The vertices A(-2,-1), B(-3, 2), C(-1, 3), and D(0, 0) form a parallelogram. The vertices A'(-1, -2), B'(2, -3), C'(3,-1),
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Asked and answered elsewhere.
brainly.com/question/9247314
You obviously don't mind using "technology" (Brainly) to answer these questions. A graphing calculator can do quadratic regression on the sequence and tell you its formula.
If you want to do it by hand, you can write the equation
.. y = ax^2 +bx +c
and substitute three of the given points. Then solve the resulting three linear equations for a, b, and c.
.. 4 = a +b +c
.. 7 = 4a +2b +c
.. 12 = 9a +3b +c
Subtracting the first equation from the other two reduces this to
.. 3 = 3a +b
.. 8 = 8a +2b
The latter can be divided by 2, so reduces to
.. 4 = 4a +b
Subtracting the first of the reduced equations from this, you have
.. 1 = a
so
.. 3 = 3*1 +b
.. 0 = b
and
.. 4 = a + b + c = 1 + 0 + c
.. 3 = c
And your equation is
.. y = x^2 +3 . . . . . . as shown previously
brainly.com/question/9247314
You obviously don't mind using "technology" (Brainly) to answer these questions. A graphing calculator can do quadratic regression on the sequence and tell you its formula.
If you want to do it by hand, you can write the equation
.. y = ax^2 +bx +c
and substitute three of the given points. Then solve the resulting three linear equations for a, b, and c.
.. 4 = a +b +c
.. 7 = 4a +2b +c
.. 12 = 9a +3b +c
Subtracting the first equation from the other two reduces this to
.. 3 = 3a +b
.. 8 = 8a +2b
The latter can be divided by 2, so reduces to
.. 4 = 4a +b
Subtracting the first of the reduced equations from this, you have
.. 1 = a
so
.. 3 = 3*1 +b
.. 0 = b
and
.. 4 = a + b + c = 1 + 0 + c
.. 3 = c
And your equation is
.. y = x^2 +3 . . . . . . as shown previously