C H A P T E R 2 Polynomial and Rational Functions

41. 2, 2 is the vertex.
42. 2, 0 is the vertex.
y ax 2 2 2
Since the graph passes through the point 1, 0, we have:
0 a1 2 2 2
2 a
y 2x 2 2 2
Section 2.1 Quadratic Functions and Models 143
f x ax 2 2 0 ax 2 2
Since the graph passes through 3, 2,
2 a3 2 2
2 a.
So, y 2x 2 2 .
43. 2, 5 is the vertex.
44. 4, 1 is the vertex.
f x ax 2 2 5
Since the graph passes through the point 0, 9, we have:
9 a0 2 2 5
4 4a
1 a
fx 1x 2 2 5 x 2 2 5
f x ax 4 2 1
Since the graph passes through 2, 3,
3 a2 4 2 1
3 4a 1
4 4a
1 a.
So, f x x 4 2 1.
45. 3, 4 is the vertex.
46. 2, 3 is the vertex.
fx ax 3 2 4
Since the graph passes through the point 1, 2, we have:
2 a1 3 2 4
2 4a
1
2 a
fx 1
2x 3 2 4
f x ax 2 2 3
Since the graph passes through 0, 2,
2 a0 2 2 3
2 4a 3
1 4a
1
4 a.
So, f x 1
4x 2 2 3.
47. 5, 12 is the vertex.
48. 2, 2 is the vertex.
fx ax 5 2 12
Since the graph passes through the point 7, 15, we have:
15 a7 5 2 12
3 4a ⇒ a 3
4
f x 3
4x 5 2 12
f x ax 2 2 2
Since the graph passes through 1, 0,
0 a1 2 2 2
0 a 2
2 a.
So, f x 2x 2 2 2.
49. is the vertex.
1 3
4 , 2 50. 2 , 3 is the vertex.
f x ax 1
4 2 3
2
Since the graph passes through the point
we have:
3
2
0 a2 1
4 2 3
2
49
16a ⇒ a 24 49
fx 24
49x 1
4 2 3
2
2, 0,
5
f x ax 5
2 2
Since the graph passes through 2, 4,
4 a2 5
2 2
3
4
4 81 3
4 a 4
19
4
4
81
4 a
19
81 a.
3
4
f x 19
81x 5
2 2
3
4
So, .