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is the image of
, and
is the preimage of
. So,
- the image of
is
- the preimage of
is
- the image of
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- the preimage of
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PLEASE HELP ME ASAP PLEASE! I KEEP GETTING SPAMMED
1 answer:
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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
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