Download Examples done in class to find equation when given the graph of a

Examples done in class to find equation when given the graph of a polynomial

Find the lowest possible degree of the polynomial

Determine the sign of the leading coefficient

Find all intercepts

Determine all linear factors and their multiplicities

Find the equation of the polynomial
y
10
9
8
7
6
5
4
3
2
1
0
-2
-1-1
-2
-3
-4
-5
-6
-7
-8
-9
-10
1
-10 -9
-8
-7
-6
-5
-4
-3
0 1
2
3
4
5
6
7
8
9
10
x
Degree 3; positive leading coefficient a3 > 0
y-intercept: y = 8
x-intercepts: x = 4, x = 2, x = 3 giving linear 3 factors (x + 4)
Equation y = a(x + 4) (x + 2) (x  3)
8 = a(0 + 4) (0 + 2) (0  3)
8 = a ( 24), solve to get a =1/ 3
1
y = (x + 4) (x + 2) (x  3)
3
y
10
9
8
7
6
5
4
3
2
1
0
-2
-1-1
-2
-3
-4
-5
-6
-7
-8
-9
-10
2
-10 -9
-8
-7
-6
-5
-4
-3
0 1
2
3
4
5
6
7
8
9
(x + 2)
(x  3)
10
x
Degree 4; positive leading coefficient a4 > 0
y-intercept: y = 8
x-intercepts: x = 4, x = 2, x = 1 giving 3 linear factors (x + 4) (x + 2)
(x  1)
Factor (x + 2) has multiplicity 2
(because it touches but does not cross axis it must have even multiplicity)
Equation y = a(x + 4) (x + 2)2 (x  1)
8 = a(0 + 4) (0 + 2) 2 (0  1)
8 = a (16), solve to get a =1/ 2
1
y=
(x + 4) (x + 2)2 (x  1)
2
Examples done in class to find equation when given the graph of a polynomial

Find the lowest possible degree of the polynomial

Determine the sign of the leading coefficient

Find all intercepts

Determine all linear factors and their multiplicities

Find the equation of the polynomial
y
10
9
8
7
6
5
4
3
2
1
0
-2
-1-1
-2
-3
-4
-5
-6
-7
-8
-9
-10
3
-10 -9
-8
-7
-6
-5
-4
-3
0 1
2
3
4
5
6
7
8
9
10
x
Degree 3; negative leading coefficient a3 < 0
y-intercept: y = 6
x-intercepts: x = 3, x = 1, x = 1 giving 3 linear factors (x + 3)
Equation y = a(x + 3) (x + 1) (x  1)
6 = a(0 + 3) (0 + 1) (0  1)
6 = a ( 3), solve to get a = 2
y = 2 (x + 3) (x + 1) (x  1)
y
10
9
8
7
6
5
4
3
2
0
1
-2
-1-1
-2
-3
-4
-5
-6
-7
-8
-9
-10
-11
-12
-13
-14
-15
-16
-17
-18
-19
-20
4
-10 -9
-8
-7
-6
-5
-4
-3
0 1
2
3
4
5
6
7
8
9
(x + 1)
(x  1)
10
x
Degree 4; negative leading coefficient a4 < 0
y-intercept: y = 9
x-intercepts: x = 4, x = 3, x = 1, x = 3 giving 4 linear factors (x + 4) (x + 3) (x  1)
Equation y = a (x + 4)(x + 3)(x  1)(x  3)
9 = a(0 + 4)(0 + 3)(0  1)(0  3)
9 = a (36), solve to get a = 1/4
1
y =  (x + 4)(x + 3)(x  1)(x  3)
4
(x  3)