Problem: The least value of the function ax2+bx+c(a>0) is:
Answer Choices:
A. βabβ
B. β2abβ
C. b2β4ac
D. 4a4acβb2β
E. none of the above.
Solution:
The graph of the function yβ‘ax2+bx+c,aξ =0, is a parabola with its axis of symmetry (the line MV extended) parallel to the y-axis.
We wish to find the coordinates of V. Let y=k be any line parallel to the x-axis which intersects the parabola at two points, say P and Q. Then the abscissa (the x-coordinate) of the midpoint of the line segment PQ is the abscissa of V. Since P and Q are the intersections of the line y=k with the parabola, the abscissas of P and Q must satisfy the equation
ax2+bx+c=k or ax2+bx+cβk=0
Its roots are:
x=β2abβ+2a1ββDΛ and xβ²=β2abββ2a1ββDΛ,
where D=b2β4ac is the discriminant.
β΄ the abscissa of V=2x+xβ²β=β2abβ, and the
ordinate of V=a(β.2abβ)2+b(β2abβ)+c=β4ab2β+c=4a4acβb2β
if a>0, the ordinate of V is the minimum value of the function ax2+bx+c and if a<0, it is the maximum value.
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