Page 137 - Algebra

11. Write the equation as a factored form f(x)=x2 –9x–22
f(x) = x2 – 11x + 2x – 22
f(x) = x(x – 11) + 2(x – 11)
f(x) = (x + 2)(x – 11)
Compare it with
f(x) = (x – b)(x – c)
x = b and x = c are two intercepts. So, in the problem, x = –2 and x = 11
12. Put (–5, 0) in y = a(x – h)2 0 = a(–5 – h)2
–5 – h = 0
h = –5
The y-intercept is 10 or (0, 10) Put (0, 16) in y = a(x – h)2
10 = a(0 – h)2
10 = ah2
10 = a(–5)2
a = 10 25
a =2 5
13. The three equations do not intersect at a single point. Therefore, there is no solution.
14. The x-intercepts of the parabola are –3 and 2. So, the equation of parabola is y = (x + 3)(x – 2)
y = x2– 2x + 3x – 6
y=x2 +x–6
Now, complete the square
211 y=x +x–6- +
44 211
y = x + x + – 6–
y = (x + ) –
44 12 25
24
The vertex of the parabola is ( 2 , 4 )
−25 The vertical distance from the point of vertex (4, –2) = –2 –( 4 )
25 −8+25 17 =–2+(4)= 4 =4
−𝑏 15. Line of symmetry x = 2𝑎
Put all the points in the equation y = ax2 + bx + c Point 1: (0, –16)
y = ax2 + bx + c
–16 = a(0) + b(0) + c
c = –16
Point 2: (1, –21)
y = ax2 + bx + c
–21 = a(1)2 + b(1) –16
–5 = a + b...........................................(1)
−1 −25
Page 136 of 177
Algebra I & II