Page 134 - Algebra

Explanation:
1. f(x) = –x2 + 10x –21
Here a < 0, so it’s a downward parabola.
Only option b has a downward parabola. So, the answer is option b.
2nd method:
a =–1 , b = 10, c = –21 The vertex is (5, 4) Put f(x) = 0
–x2 +10x–21=0
–x2 + 7x + 3x –21 –x(x –
y=–x2 +10x–21 y = –21
–21)
2. Here a < 0, so the function has maxima
a = – 1 , b = 5, c = – 12
2
−23 =4
(x – 7) (
x-intercepts are (7, 0) and (3, 0)
Put x = 0,
y –intercept is (0,
Now, plot these points on the graph.
x= – =
x =–
𝑏 2𝑎
7) + 3(x – 7) = 0
55
−2 2
–
x + 3) = 0
x = 7 or 3
x =–
Put f(5) in –x2 + 5x –12
= – = –
– 12 – 12
(5)2 + 5(5) 22
25 + 25 42
−25+50−48 4
=
3. Here a > 0, so the function has minima x2 + 10x + 20
a = 1, b = 10, c = 20
𝑏 2𝑎
x = –5
Put f(-5) in x2 + 10x + 20 = (-5) + 20
=–5
2 + 10(-5)
= 25 – 50 + 20
4. Here a > 0, so the function has minima
Page 133 of 177
Algebra I & II