the graph of y=x^2-6x+5 could be
Answers
Answer:
Step-by-step explanation:
If we want to take a more algebraic/analytical approach:
Determine the function's zeros:
We can find where the function crosses the
x
-axis by setting the function equal to
0
.
x
2
−
6
x
+
5
=
0
Factor this by looking for two numbers that add up to
−
6
and multiply to
5
. In this case, these are
−
5
and
−
1
.
(
x
−
5
)
(
x
−
1
)
=
0
x
=
5
x
=
1
We know the graph of the parabola will cross the
x
-axis at these two points. We can mark them on a graph:
graph{((x-5)^2+y^2-1/30)((x-1)^2+y^2-1/30)=0 [-7.33, 15.18, -6.24, 7.01]}
Determine the function's y-intercept:
The
y
-intercept will occur when
x
=
0
:
y
=
0
2
−
6
(
0
)
+
5
=
5
There is a
y
-intercept at
(
0
,
5
)
, which we can mark on our graph.
graph{((x-5)^2+y^2-1/30)((x-1)^2+y^2-1/30)(x^2+(y-5)^2-1/30)=0 [-7.33, 15.18, -6.24, 7.01]}
Determining the vertex:
The
x
-coordinate of the vertex of a parabola in the form
a
x
2
+
b
x
+
c
can be found through the formula
x
vertex
=
−
b
2
a
(This is also on the vertical line where the parabola's axis of symmetry lies.)
For
x
2
−
6
x
+
5
, we see that
a
=
1
and
b
=
−
6
, so
x
vertex
=
−
−
6
2
⋅
1
=
6
2
=
3
The
y
-coordinate of the vertex can be found through plugging in
x
=
3
, which is the
x
-coordinate of the vertex, into the parabola's equation:
y
=
3
2
−
6
(
3
)
+
5
=
9
−
18
+
5
=
−
4
So we know
(
3
,
−
4
)
is the vertex of the parabola.
graph{((x-3)^2+(y+4)^2-1/20)((x-5)^2+y^2-1/30)((x-1)^2+y^2-1/30)(x^2+(y-5)^2-1/30)=0 [-7.33, 15.18, -6.24, 7.01]}
Using the symmetric properties of a parabola, since we know
(
0
,
5
)
is a point,
3
units over from the vertex, we know
(
6
,
5
)
will also be on the parabola since it is
3
units over in the other direction:
graph{((x-6)^2+(y-5)^2-1/30)((x-3)^2+(y+4)^2-1/20)((x-5)^2+y^2-1/30)((x-1)^2+y^2-1/30)(x^2+(y-5)^2-1/30)=0 [-7.33, 15.18, -6.24, 7.01]}
Using these points, we can draw the parabola quite well:
graph{x^2-6x+5 [-8.36, 14.15, -5.5, 7.75]}
hope it helps !!
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