given a parabola in standard form
y
=
a
x
2
+
b
x
+
c
then the x-coordinate of the vertex can be found using
∙
x
x
vertex
=
−
b
2
a
y
=
x
2
−
6
x
+
5
is in standard form
with
a
=
1
,
b
=
−
6
,
c
=
5
⇒
x
vertex
=
−
−
6
2
=
3
substitute this value into equation for y-coordinate
y
vertex
=
3
2
−
6
(
3
)
+
5
=
−
4
⇒
vertex
=
(
3
,
−
4
)
the axis of symmetry is vertical and passes through the
vertex with equation
x
=
3
to find x-intercepts let y = 0
⇒
x
2
−
6
x
+
5
=
0
the factors of + 5 which sum to - 6 are - 1 and - 5
⇒
(
x
−
1
)
(
x
−
5
)
=
0
equate each factor to zero and solve for x
x
−
1
=
0
⇒
x
=
1
x
−
5
=
0
⇒
x
=
5
⇒
x
=
1
and
x
=
5
←
x-intercepts
graph{(y-x^2+6x-5)(y-1000x+3000)=0 [-10, 10, -5, 5]}