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

Skip to content# Find an equation in standard form of the parabola passing through these points: (0,-,-,0)

##
This Post Has 3 Comments

### Leave a Reply

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

The standard form of a parabola is y=ax²+bx+c

use the three given points to find the three unknown constants a, b, and c:

-2=a+b+c1

-2=4a+2b+c 2

-4=9a+3b+c3

equation 2 minus equation 1: 3a+b=04

equation 3 minus equation 2: 5a+b=-25

equation 5 minus equation 4: 2a=-2, so a=-1

plug a=-1 in equation 4: -3+b=0, so b=3

Plug a=-1, b=3 in equation 1: -2=-1+3+c, so c=-4

the parabola is y=-x²+3x-4

double check: when x=1, y=-1+3-4=-2

when x=2, y=-4+6-4=-2

when x=3, y=-9+9-4=-4

Yes.

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.

y = -(x - 2.5)² + 0.25

[tex]Find an equation in standard form of the parabola passing through these points: (0,-,-,0)[/tex]

Help. Mnfded tyyyugffghjjbvffyuivvcd