A quadratic having zeros 1/2 and - 5/2 is
y = ƒ(x) = a( x - 1/2 )( x + 5/2 ), ..... a > 0
.... ƒ(x) = (a/4) * ( 2x - 1 )( 2x + 5 )
.... ƒ(x) = (a/4) * ( 4x² + 8x - 5 ) .....................(1)
....................................................................................................................................................
Diff ... w.r.t. x
ƒ'(x) = (a/4) * ( 8x + 8 ) ......... ƒ'(x) = 0 => x = -1
ƒ''(x) = (a/4)*8 = 2a .............. ƒ''(-1) = 2a > 0
Hence, ... x = -1 ... gives minimum of ƒ(x).
.................................................................................................................................................
Since, this min. is = -9, .... from (1),
ƒ(-1) = -9
(a/4)*[ 4(-1)² + 8(-1) - 5 ] = -9
(a/4)*(- 9 ) = -9
a = 4 .................................................. (2)
...........................................................................................................................................
From (1) and (2), the req'd quadratic is
ƒ(x) = 4x² + 8x - 5 ...................................Ans.
..............................................................................................................................................
Happy To Help !
................................................................................................................................................
Y(x) = ax^2 + bx + c
a = 4.00000
b = 8.00000
c = -5.00000
or
V(x) = a(x-h)^2 + k
h = -1.00000
k = -9.00000
Y(x) = a ( x-x1 )( x-x2 )
a = 4
x1 = 0.50000
x2 = -2.50000
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Verified answer
A quadratic having zeros 1/2 and - 5/2 is
y = ƒ(x) = a( x - 1/2 )( x + 5/2 ), ..... a > 0
.... ƒ(x) = (a/4) * ( 2x - 1 )( 2x + 5 )
.... ƒ(x) = (a/4) * ( 4x² + 8x - 5 ) .....................(1)
....................................................................................................................................................
Diff ... w.r.t. x
ƒ'(x) = (a/4) * ( 8x + 8 ) ......... ƒ'(x) = 0 => x = -1
ƒ''(x) = (a/4)*8 = 2a .............. ƒ''(-1) = 2a > 0
Hence, ... x = -1 ... gives minimum of ƒ(x).
.................................................................................................................................................
Since, this min. is = -9, .... from (1),
ƒ(-1) = -9
(a/4)*[ 4(-1)² + 8(-1) - 5 ] = -9
(a/4)*(- 9 ) = -9
a = 4 .................................................. (2)
...........................................................................................................................................
From (1) and (2), the req'd quadratic is
ƒ(x) = 4x² + 8x - 5 ...................................Ans.
..............................................................................................................................................
Happy To Help !
................................................................................................................................................
Y(x) = ax^2 + bx + c
a = 4.00000
b = 8.00000
c = -5.00000
or
V(x) = a(x-h)^2 + k
a = 4.00000
h = -1.00000
k = -9.00000
or
Y(x) = a ( x-x1 )( x-x2 )
a = 4
x1 = 0.50000
x2 = -2.50000