How to Find the Maximum or Minimum Value of a Quadratic Function Easily
Set up the function in general form., Determine the direction of the graph., Calculate -b/2a., Find the corresponding f(x) value., Report your results.
Step-by-Step Guide
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Step 1: Set up the function in general form.
A quadratic function is one that has an x2{\displaystyle x^{2}} term.
It may or may not contain an x{\displaystyle x} term without an exponent.
There will be no exponents larger than
2.
The general form is f(x)=ax2+bx+c{\displaystyle f(x)=ax^{2}+bx+c}.
If necessary, combine similar terms and rearrange to set the function in this general form.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} -
Step 2: Determine the direction of the graph.
A quadratic function results in the graph of a parabola.
The parabola either opens upward or downward.
If a{\displaystyle a}, the coefficient of the x2{\displaystyle x^{2}} term, 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)=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. , The value of −b2a{\displaystyle
-{\frac {b}{2a}}} tells you the x{\displaystyle x} value of the vertex of the parabola.
When the quadratic function is written in its general form of ax2+bx+c{\displaystyle ax^{2}+bx+c}, use the coefficients of the x{\displaystyle x} and x2{\displaystyle x^{2}} terms as follows:
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} , Insert the value of x that you just calculated into the function to find the corresponding value of f(x).
This will be the minimum or maximum of the function.
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(x)=(−5)2+10(−5)−1{\displaystyle f(x)=(-5)^{2}+10(-5)-1} f(x)=25−50−1{\displaystyle f(x)=25-50-1} f(x)=−26{\displaystyle f(x)=-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(x)=−3(1)2+6(1)−4{\displaystyle f(x)=-3(1)^{2}+6(1)-4} f(x)=−3+6−4{\displaystyle f(x)=-3+6-4} f(x)=−1{\displaystyle f(x)=-1} , Review the question you have been asked.
If you are asked for the coordinates of the vertex, you need to report both the x{\displaystyle x} and y{\displaystyle y} (or f(x){\displaystyle f(x)}) values.
If you are only asked for the maximum or minimum, you only need to report the y{\displaystyle y} (or f(x){\displaystyle f(x)}) value.
Refer back to the value of the a{\displaystyle a} coefficient to be sure if you have a maximum or a minimum.
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}. -
Step 3: Calculate -b/2a.
-
Step 4: Find the corresponding f(x) value.
-
Step 5: Report your results.
Detailed Guide
A quadratic function is one that has an x2{\displaystyle x^{2}} term.
It may or may not contain an x{\displaystyle x} term without an exponent.
There will be no exponents larger than
2.
The general form is f(x)=ax2+bx+c{\displaystyle f(x)=ax^{2}+bx+c}.
If necessary, combine similar terms and rearrange to set the function in this general form.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}
A quadratic function results in the graph of a parabola.
The parabola either opens upward or downward.
If a{\displaystyle a}, the coefficient of the x2{\displaystyle x^{2}} term, 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)=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. , The value of −b2a{\displaystyle
-{\frac {b}{2a}}} tells you the x{\displaystyle x} value of the vertex of the parabola.
When the quadratic function is written in its general form of ax2+bx+c{\displaystyle ax^{2}+bx+c}, use the coefficients of the x{\displaystyle x} and x2{\displaystyle x^{2}} terms as follows:
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} , Insert the value of x that you just calculated into the function to find the corresponding value of f(x).
This will be the minimum or maximum of the function.
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(x)=(−5)2+10(−5)−1{\displaystyle f(x)=(-5)^{2}+10(-5)-1} f(x)=25−50−1{\displaystyle f(x)=25-50-1} f(x)=−26{\displaystyle f(x)=-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(x)=−3(1)2+6(1)−4{\displaystyle f(x)=-3(1)^{2}+6(1)-4} f(x)=−3+6−4{\displaystyle f(x)=-3+6-4} f(x)=−1{\displaystyle f(x)=-1} , Review the question you have been asked.
If you are asked for the coordinates of the vertex, you need to report both the x{\displaystyle x} and y{\displaystyle y} (or f(x){\displaystyle f(x)}) values.
If you are only asked for the maximum or minimum, you only need to report the y{\displaystyle y} (or f(x){\displaystyle f(x)}) value.
Refer back to the value of the a{\displaystyle a} coefficient to be sure if you have a maximum or a minimum.
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}.
About the Author
Jacob Ortiz
Specializes in breaking down complex crafts topics into simple steps.
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