Asked by Sandra
A parabola has x-intercepts at 3 and 7, a y-intercept at -21, and (5,4) for its vertex. Write the parabola’s equation
Answers
Answered by
Damon
Sketch those points on a graph.
It is an upside down parabola (sheds water)
The axis of symmetry is x = 5
when x = 5, y = 4
so
for starters
y-4 = c (x-5)^2
That is the same both sides of x = 5
and y = 4 when x = 5
We need c
put in one of those other points
for example (0,-21)
-21 -4 = c (-5)^2
-25 = c (25)
c = -1
so
y-4 = -(x-5)^2
or
y = 4 - (x-5)^2
check with (3,0)
? = 4 -(-2)^2
? = 4 - 4
0 sure enough
It is an upside down parabola (sheds water)
The axis of symmetry is x = 5
when x = 5, y = 4
so
for starters
y-4 = c (x-5)^2
That is the same both sides of x = 5
and y = 4 when x = 5
We need c
put in one of those other points
for example (0,-21)
-21 -4 = c (-5)^2
-25 = c (25)
c = -1
so
y-4 = -(x-5)^2
or
y = 4 - (x-5)^2
check with (3,0)
? = 4 -(-2)^2
? = 4 - 4
0 sure enough
Answered by
henry2,
Given: r1(3, 0), r2(7, 0). y-int.(0, -21). V(5, 4).
Y = a(x-h)^2 + k.
a(3 - 5)^2 + 4 = 0,
4a = -4,
a = -1.
Vertex form: Y = -1(x - 5)^2 + 4.
Y = -1(x^2 - 10x + 25) + 4,
Y = -x^2 + 10x - 21(Standard form).
Y = a(x-h)^2 + k.
a(3 - 5)^2 + 4 = 0,
4a = -4,
a = -1.
Vertex form: Y = -1(x - 5)^2 + 4.
Y = -1(x^2 - 10x + 25) + 4,
Y = -x^2 + 10x - 21(Standard form).
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