1.It is given that (2,5) and (-1,11) are two points on the graph of the quadratic function y=ax^2+bx+1 .
a)Find the values of a and b .
b)Find the maximum or minimum value of y and the corresponding value of x.
c)write down the vertex and the axis of symmetry of the graph of the function.
thx..
MATHS....
2007-11-04 9:40 pm
回答 (1)
2007-11-04 10:00 pm
✔ 最佳答案
(a) 5 = a(2)^2 + 2b + 14a+2b = 4
2a+b = 2 ...(1)
11 = a(-1)^2 - b + 1
a - b = 10 ...(2)
(1)+ (2): 3a = 12
a = 4
b = -6
(b) y = 4x^2 - 6x + 1
y = 4(x^2 - 3/2 x) + 1
y= 4(x^2 - 3/2 x + (3/4)^2) + 1 - 4(3/4)^2
y = 4(x-3/4)^2 + 1 - 9/4
y = 4(x-3/4)^2 - 5/4
(a) 5 = a(2)^2 + 2b + 1
4a+2b = 4
2a+b = 2 ...(1)
11 = a(-1)^2 - b + 1
a - b = 10 ...(2)
(1)+ (2): 3a = 12
a = 4
b = -6
(b) y = 4x^2 - 6x + 1
y = 4(x^2 - 3/2 x) + 1
y= 4(x^2 - 3/2 x + (3/4)^2) + 1 - 4(3/4)^2
y = 4(x-3/4)^2 + 1 - 9/4
y = 4(x-3/4)^2 - 5/4
Hence the minimum value of y is -5/4 when x = 3/4
(c) The vertex of the graph is (3/4, -5/4)
The axis of symmetry is x = 3/4
Hence the minimum value of y is -5/4 when x = 3/4
(c) The vertex of the graph is (3/4, -5/4)
The axis of symmetry is x = 3/4
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