quadratics

[red and white kop!]

the equation of a curve is y=a(x^2)-2bx+c, where a, b and c are constants with a>0
given that the verte of the curve lies on the line y=x, find an expression for c in terms of a and b. show that in this case, whatever the value of b, c>= -1/4a

so i got c= b(b+1)/a or something similar, but the problem is i am totally lost as to where this equation fits in with c>= -1/4a
can someone explain this to me

 

[Deleted member 4993]

the equation of a curve is y=a(x^2)-2bx+c, where a, b and c are constants with a>0
given that the verte of the curve lies on the line y=x, find an expression for c in terms of a and b. show that in this case, whatever the value of b, c>= -1/4a

so i got c= b(b+1)/a or something similar, but the problem is i am totally lost as to where this equation fits in with c>= -1/4a

Is it

c ? - 1/4 * a (= -a/4) or

c ? - 1/(4 * a)

By this time you ought to know that those two are totally different. Use grouping symbols to avoid confusion (and wastage of time).

can someone explain this to me

 

[red and white kop!]

-1/(4a)

 

[Deleted member 4993]

the equation of a curve is y=a(x^2)-2bx+c, where a, b and c are constants with a>0
given that the verte of the curve lies on the line y=x, find an expression for c in terms of a and b. show that in this case, whatever the value of b, c>= -1/4a

so i got c= b(b+1)/a or something similar, but the problem is i am totally lost as to where this equation fits in with c>= -1/4a
can someone explain this to me

Use the method you learned in completing-square.

Spoiler: :1d3j64xs

c = b*(b+1)/a = [(b+1/2)[sup:1d3j64xs]2[/sup:1d3j64xs]-1/4]/a = (b+1/2)[sup:1d3j64xs]2[/sup:1d3j64xs]/a - 1/(4a) ..... continue....[/spoiler:1d3j64xs]