Page 138 - Algebra

Point 3: (–1, –9)
y = ax2 + bx + c
–9 = a(–1)2 + b(–1) –16 –9 = a –b – 16
a –b = 7............................................(2)
Subtract (1) and (2) a + b = –5
a– b = 7
2b = –12
b = –6 a– b = 7 a + 6= 7 a =1
Line of symmetry x =
−(−6) 2(1) = 3
16. You can find the point of intersection by equating the two equations. f(x) = g(x)
x2 –10=–x2 +8 2x2 = 18
x2 = 9
x = ±3
f(x)=x2 –10
f(3)=32 –10=9–10=–1 f(–3)=(–3)2 –10=9–10=–1 So, the points are (3, –1), (–3, –1) So, b = –1
Page 137 of 177
Algebra I & II