draw the graph of polynomial x² +2x-15 & find the zeros
Answers
Part 1.
Let, y = x² + 2x - 15
Now, y = x² + 2x - 15
or, y = (x + 1)² - 16
or, (x + 1)² = y + 16 ..... (1)
This is an equation of a parabola whose vertex is at (- 1, - 16) and its axis x = - 1 is parallel to the y-axis.
Putting y = 0 in (1), we get
(x + 1)² = 16
or, (x + 1)² - 4² = 0
or, (x + 5) (x - 3) = 0
i.e., x = - 5, 3
So the parabola (1) intersects the x-axis at (- 5, 0) and (3, 0)
To draw the graph:
• Take the vertex (- 1, - 16) on the graph paper
• Take (- 5, 0) and (3, 0) on the x-axis
• Join (- 1, -16) with (- 5, 0) and (3, 0) with curves
• Stretch the two lines a little on upper side
# Take a look at the attached picture.
Part 2.
Now, x² + 2x - 15
= x² + 5x - 3x - 15
= x (x + 5) - 3 (x + 5)
= (x + 5) (x - 3)
This is the required factorization.
Thus the required zeroes are (- 5) and 3
Answer:
Given -
y
=
x
2
−
2
x
−
15
It can be graphed by plotting the following points
vertex, y-intercept, and x-intercepts
Vertex
x
=
−
b
2
a
)
=
−
(
−
2
)
2
×
1
=
1
At
x
=
1
y
=
1
2
−
2
(
1
)
−
15
=
1
−
2
−
15
=
−
16
Vertex
(
1
,
−
16
)
Y-Intercept
At
x
=
0
y
=
0
2
−
2
(
0
)
−
15
=
−
15
Y-Intercept
(
0
,
−
15
)
Its x-intercepts are
At
y
=
0
x
2
−
2
x
−
15
=
0
x
2
−
5
x
+
3
x
−
15
=
0
x
(
x
−
5
)
+
3
(
x
−
5
)
=
0
(
x
+
3
)
(
x
−
5
)
=
0
x
=
−
3
x
=
5
The two x- intercepts are
(
−
3
,
0
)
(
5
,
0
)
Plot the points
(
1
,
−
16
)
;
(
0
,
−
15
)
;
(
−
3
,
0
)
(
5
,
0
)
You will get the curve
Look at the graph