plot y=-x²+4x-4 to obtain a parabola opening downward
Answers
Answer:
hey buddy here it is
Step-by-step explanation:
Vertex:
(
2
,
0
)
, y intercept:
(
0
,
4
)
, x intercept:
(
2
,
0
)
, symmetry line :
x
=
2
, additional points :
(
0
,
4
)
,
(
4
,
4
)
and
(
1.5
,
0.25
)
,
(
2.5
,
0.25
)
Explanation:
y
=
x
2
−
4
x
+
4
or
y
=
(
x
−
2
)
2
+
0
This is vertex form of equation ,
y
=
a
(
x
−
h
)
2
+
k
;
(
h
,
k
)
being vertex , here
h
=
2
,
k
=
0
,
a
=
1
Since
a
is positive, parabola opens upward.
Therefore vertex is at
(
h
,
k
)
or
(
2
,
0
)
Axis of symmetry is
x
=
h
or
x
=
2
;
x-intercept is found by putting
y
=
0
in the equation
y
=
(
x
−
2
)
2
or
(
x
−
2
)
2
=
0
or
or
x
=
2
or
(
2
,
0
)
or
y-intercept is found by putting
x
=
0
in the equation
y
=
(
x
−
2
)
2
or
y
=
(
0
−
2
)
2
or
y
=
4
or
(
0
,
4
)
. Graph points:
Distance of vertex from directrix is
d
=
1
4
|
a
|
or
1
4
The length of a parabola's latus rectum is
4
d
=
1
, where "d" is the
distance from the focus to the vertex. Ends of the latus rectum
are
(
1.5
,
0.25
)
and
(
2.5
,
0.25
)
Additional point is
(
0
,
4
)
and
(
4
,
4
)
graph{(x-2)^2 [-10, 10, -5, 5]}[Ans]