For example, suppose you start with f(x)=3x+2x−x2+3x2+4{\displaystyle f(x)=3x+2x-x^{2}+3x^{2}+4}. Combine the x2{\displaystyle x^{2}} terms and the x{\displaystyle x} terms to get the following in general form: f(x)=2x2+5x+4{\displaystyle f(x)=2x^{2}+5x+4}
For f(x)=2x2+4x−6{\displaystyle f(x)=2x^{2}+4x-6}, a=2{\displaystyle a=2} so the parabola opens upward. For f(x)=−3x2+2x+8{\displaystyle f(x)=-3x^{2}+2x+8}, a=−3{\displaystyle a=-3} so the parabola opens downward. For f(x)=x2+6{\displaystyle f(x)=x^{2}+6}, a=1{\displaystyle a=1} so the parabola opens upward. If the parabola opens upward, you will be finding its minimum value. If the parabola opens downward, you will find its maximum value.
For a function f(x)=x2+10x−1{\displaystyle f(x)=x^{2}+10x-1}, a=1{\displaystyle a=1} and b=10{\displaystyle b=10}. Therefore, find the x-value of the vertex as: x=−b2a{\displaystyle x=-{\frac {b}{2a}}} x=−10(2)(1){\displaystyle x=-{\frac {10}{(2)(1)}}} x=−102{\displaystyle x=-{\frac {10}{2}}} x=−5{\displaystyle x=-5} As a second example, consider the function f(x)=−3x2+6x−4{\displaystyle f(x)=-3x^{2}+6x-4}. In this example, a=−3{\displaystyle a=-3} and b=6{\displaystyle b=6}. Therefore, find the x-value of the vertex as: x=−b2a{\displaystyle x=-{\frac {b}{2a}}} x=−6(2)(−3){\displaystyle x=-{\frac {6}{(2)(-3)}}} x=−6−6{\displaystyle x=-{\frac {6}{-6}}} x=−(−1){\displaystyle x=-(-1)} x=1{\displaystyle x=1}
For the first example above, f(x)=x2+10x−1{\displaystyle f(x)=x^{2}+10x-1}, you calculated the x-value for the vertex to be x=−5{\displaystyle x=-5}. Enter −5{\displaystyle -5} in place of x{\displaystyle x} in the function to find the maximum value: f(x)=x2+10x−1{\displaystyle f(x)=x^{2}+10x-1} f(−5)=(−5)2+10(−5)−1{\displaystyle f(-5)=(-5)^{2}+10(-5)-1} f(−5)=25−50−1{\displaystyle f(-5)=25-50-1} f(−5)=−26{\displaystyle f(-5)=-26} For the second example above, f(x)=−3x2+6x−4{\displaystyle f(x)=-3x^{2}+6x-4}, you found the vertex to be at x=1{\displaystyle x=1}. Insert 1{\displaystyle 1} in place of x{\displaystyle x} in the function to find the maximum value: f(x)=−3x2+6x−4{\displaystyle f(x)=-3x^{2}+6x-4} f(1)=−3(1)2+6(1)−4{\displaystyle f(1)=-3(1)^{2}+6(1)-4} f(1)=−3+6−4{\displaystyle f(1)=-3+6-4} f(1)=−1{\displaystyle f(1)=-1}
For the first example, f(x)=x2+10x−1{\displaystyle f(x)=x^{2}+10x-1}, the value of a{\displaystyle a} is positive, so you will be reporting the minimum value. The vertex is at (−5,−26){\displaystyle (-5,-26)}, and the minimum value is −26{\displaystyle -26}. For the second example, f(x)=−3x2+6x−4{\displaystyle f(x)=-3x^{2}+6x-4}, the value of a{\displaystyle a} is negative, so you will be reporting the maximum value. The vertex is at (1,−1){\displaystyle (1,-1)}, and the maximum value is −1{\displaystyle -1}.
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.
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.
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.
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).
Begin with the sample function f(x)=2x2−4x+1{\displaystyle f(x)=2x^{2}-4x+1}.
For the sample function f(x)=2x2−4x+1{\displaystyle f(x)=2x^{2}-4x+1}, find the derivative as: f′(x)=4x−4{\displaystyle f^{\prime }(x)=4x-4}
f′(x)=4x−4{\displaystyle f^{\prime }(x)=4x-4} 0=4x−4{\displaystyle 0=4x-4}
0=4x−4{\displaystyle 0=4x-4} 4=4x{\displaystyle 4=4x} 1=x{\displaystyle 1=x}
For the function f(x)=2x2−4x+1{\displaystyle f(x)=2x^{2}-4x+1} at x=1{\displaystyle x=1}, f(1)=2(1)2−4(1)+1{\displaystyle f(1)=2(1)^{2}-4(1)+1} f(1)=2−4+1{\displaystyle f(1)=2-4+1} f(1)=−1{\displaystyle f(1)=-1}
