Vertex of Quadratic Function

[mathdad]

The x-coordinate of the vertex of f(x) = ax^2 + bx + c is -b/(2a). What is the y-coordinate of the vertex?

My Answer:

The y-coordinate is f(-b/2a). So, the vertex is
[(-b/2a), f(-b/2a)]. Is this the only way to express the vertex? What relation is there between the point given above and (h, k)?

 

[MarkFL]

You could go head and write:

[MATH]f\left(-\frac{b}{2a}\right)=a\left(-\frac{b}{2a}\right)^2+b\left(-\frac{b}{2a}\right)+c=?[/MATH]

 

[mathdad]

(-b/2)^2 - (b^2)/(2a) + c

(b^2)/4 - (b^2)/(2a) + c

 

[MarkFL]

Your first term is incorrect... you have the square of \(a\) in the denominator, but only \(a\) in the numerator, so you cannot cancel completely.

 

[mathdad]

Your first term is incorrect... you have the square of \(a\) in the denominator, but only \(a\) in the numerator, so you cannot cancel completely.

I see my error and will try again.