Question

Find an equation in standard form of the parabola passing through these points: (0,-6),(-1,-12),(3,0)

Answer 1

One form of the equation of a parabola is
y = ax² + bx + c

The curve passes through (0,-6), (-1,-12) and (3,0). Therefore
c = - 6                       (1)
a - b + c = -12            (2)
9a + 3b + c = 0         (3)

Substitute (1) into (2) and into (3).
a - b -6 = -12
a - b = -6                  (4)
9a + 3b - 6 = 0
9a + 3b = 6              (5)

Substitute a = b - 6 from (4) into (5).
9(b - 6) + 3b = 6
12b - 54 = 6
12b = 60
b = 5
a = b - 6 = -1

The equation is
y = -x² + 5x - 6 

Let us use completing the square to write the equation in standard form for a parabola.
y = -[x² - 5x] - 6
   = -[ (x - 2.5)² - 2.5²] - 6
   = -(x - 2.5)² + 6.25 - 6

y = -(x - 2.5)² + 0.25
This is the standdard form of the equation for the parabola.
The vertex us at (2.5, 0.25).
The axis of symmetry is x = 2.5
Because the leading coefficient is -1 (negative), the curve opens downward.
The graph is shown below.

Answer: y = -(x - 2.5)² + 0.25