The standard form of a general quadratic function, which can also be called the vertex form, looks like this:  f(x)=a(x−h)2+k{\displaystyle f(x)=a(x-h)^{2}+k} If your function is already given to you in this form, you just need to recognize the variables a{\displaystyle a}, h{\displaystyle h} and k{\displaystyle k}. If your function begins in the general form f(x)=ax2+bx+c{\displaystyle f(x)=ax^{2}+bx+c}, you will need to complete the square to rewrite it in vertex form. To review how to complete the square, see Complete the Square. Just as with a quadratic function written in its general form, you can tell the direction of the parabola by looking at the coefficient a{\displaystyle a}. If a{\displaystyle a} in this standard form is positive, then the parabola opens upward. If a{\displaystyle a} is negative, then the parabola opens downward. Look at the following examples:  For f(x)=2(x+1)2−4{\displaystyle f(x)=2(x+1)^{2}-4}, a=2{\displaystyle a=2}, which is positive, so the parabola opens upward. For f(x)=−3(x−2)2+2{\displaystyle f(x)=-3(x-2)^{2}+2}, a=−3{\displaystyle a=-3}, which is negative, so the parabola opens downward. If the parabola opens upward, you will be finding its minimum value. If the parabola opens downward, you will find its maximum value. When the function is written in standard form, finding the minimum or maximum value is as simple as stating the value of the variable k{\displaystyle k}. For the two example functions given above, these values are:  For f(x)=2(x+1)2−4{\displaystyle f(x)=2(x+1)^{2}-4}, k=−4{\displaystyle k=-4}. This is the minimum value of the function because this parabola opens upward. For f(x)=−3(x−2)2+2{\displaystyle f(x)=-3(x-2)^{2}+2}, k=2{\displaystyle k=2}. This is the maximum value of the function, because this parabola opens downward. If you are asked for the coordinates of the minimum or maximum value, the point will be (h,k){\displaystyle (h,k)}. Note, however, that in the standard form of the equation, the term inside the parentheses is (x−h){\displaystyle (x-h)}, so you need the opposite sign of the number that follows the x{\displaystyle x}.  For f(x)=2(x+1)2−4{\displaystyle f(x)=2(x+1)^{2}-4}, the term inside the parentheses is (x+1), which can be rewritten as (x-(-1)). Thus, h=−1{\displaystyle h=-1}. Therefore, the coordinates of the vertex for this function are (−1,−4){\displaystyle (-1,-4)}. For f(x)=−3(x−2)2+2{\displaystyle f(x)=-3(x-2)^{2}+2}, the term inside the parentheses is (x-2). Therefore, h=2{\displaystyle h=2}. The coordinates of the vertex are (2, 2).
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