Question 1128589
The point {{{A}}} lies on the parabola 
{{{y=x^2 - 6x +24}}}

The point {{{B}}} lies on the parabola {{{y= 6x-x^2}}} and is {{{vertically}}}{{{ below }}}the point{{{ A}}}.

solution:

{{{y=x^2 - 6x +24}}} -> parabola is {{{opening}}}{{{ up}}}, the point {{{A}}} in  vertex (minimum)

{{{y= -x^2+6x}}} -> parabola is {{{opening}}}{{{ down}}}, the point {{{B}}} in  vertex (maximum)


writ both equations in vertex form:

{{{y=(x^2 - 6x) +24}}}

{{{y=(x^2 - 6x+b^2) -b^2+24}}}

{{{y=(x^2 - 6x+3^2) -3^2+24}}}

{{{y=(x - 3)^2 -9+24}}}

{{{y=(x - 3)^2 +15}}}

=> vertex is at ({{{3}}},{{{15}}})=>the point {{{A }}}


{{{y= -x^2+6x}}}

{{{y= -(x^2-6x+b^2)-1(-b^2)}}}

{{{y= -(x^2-6x+3^2)-1(-3^2)}}}

{{{y= -(x-3)^2+9}}}

=> vertex is at ({{{3}}},{{{9}}})=>the point {{{B}}}

since both points have same {{{x}}} coordinate, the  point {{{B}}} lies vertically below the point {{{A}}} because {{{y}}} coordinate of the point {{{A}}} is greater than {{{y}}} coordinate of the point {{{B}}}



a) Show this information on a neat diagram

{{{drawing ( 600, 600, -20, 20, -20, 20,
circle(3,15,.12),circle(3,9,.12),

locate(3,15,A(3,15)), locate(3,9,B(3,9)),

line(3,15,3,9),

graph( 600, 600, -20, 20, -20, 20, (x - 3)^2 +15, -(x-3)^2+9)) }}}



b) Find the least value of the length of line AB 

the length of line {{{AB=15-9=6}}}