How to Find the Peak of a Quadratic Equation

The vertex of a quadratic equation or parabp is the highest or lowest point of that equation. It lies on the plane of symmetry of the entire parabp; any point to the left of the parabp is also a full mirror image of the point to the right. If you want to find the vertex of a quadratic equation, you can use the vertex formula, or the square’s complement.

Table of Contents

Using Peak Finding Formula

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Determine the values of a, b, and c. In the quadratic equation, the coefficient of x ² = a , the coefficient of x = b, and the constant = c . Suppose we have the following equation: y = x ² + 9x + 18 . In this example, a = 1, b = 9, and c = 18. ^{[1] X Research Source}

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Use the vertex formula to find the x value of the parabp vertex. The vertex is also the axis of symmetry of the equation. The formula for finding the x value of the vertex of a quadratic equation is x = -b/2a . Substitute the corresponding values to find x :

x=-b/2a
x=-(9)/(2)(1)
x=-9/2

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Substitute x into the original equation to find y. Once you know the value of x, just plug it into the original formula and you’ll get y. You can think of the formula for the vertex of a quadratic function as (x, y) = [(-b/2a), f(-b/2a)] . This means that to find the y value, you have to find the x value based on the given formula and then plug it into the equation. Here’s how:

y = x ² + 9x + 18
y = (-9/2) ² + 9(-9/2) +18
y = 81/4 -81/2 + 18
y = 81/4 -162/4 + 72/4
y = (81 – 162 + 72)/4
y = -9/4

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Write the values of x and y in coordinate order. Now that you know x = -9/2, and y = -9/4, just write them in order of coordinates: (-9/2, -9/4). The top of this quadratic equation is (-9/2, -9/4). If you graph this parabpa, this will be the base of the parabpa, because the coefficient of x ² is positive.

Square’s complement

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Write down the equation. The square’s complement is another way to find the vertex of a quadratic equation. With this method, you can immediately find out the coordinates of x and y instead of finding x first and then substituting x into the original equation to find y. Suppose we have the following quadratic equation: x ² + 4x + 1 = 0. ^{[2] X Research Source}

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Divide each term by the coefficient of x ² . In this example, the coefficient of x ² is 1, so you can skip this step.

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Move the constant to the right side of the equation. A constant is a term that does not change. In this example, the constant is equal to “1”. Move 1 to the other side of the equation by subtracting both sides by 1. The procedure is as follows: ^{[3] X Research Source}

x ² + 4x + 1 = 0
x ² + 4x + 1 -1 = 0 – 1
x ² + 4x = – 1

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Compensate for the square of the left side of the equation. To do this, simply find (b/2) ² and add the result to both sides of the equation. Substitute “4” for b , because “4x” is the b term of this equation.

(4/2) ² = 2 ² = 4. Now add 4 to both sides of the equation, we have:
- x ² + 4x + 4 = -1 + 4
- x ² + 4x + 4 = 3

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Factorize the left side of the equation. You can see that x ² + 4x + 4 is a perfect square. It can be rewritten as (x + 2) ² = 3

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Use this format to find the x and y coordinates. You can find the x coordinate by setting (x + 2) ² equal to 0. When (x + 2) ² = 0, x will equal -2, so your x coordinate is -2. The y coordinate is the constant on the other side of the equation. So y = 3. You can also make the shortcut by taking the opposite sign of the number inside the brackets to get the x coordinate. So the vertex of the equation x ² + 4x + 1 = (-2, 3)

Advice

Correctly identify a, b, and c.
The operations must be in order to get the correct result.

Warning

Check your results!
Make sure a, b, and c are correct – otherwise, the answer will be wrong.
Don’t worry too much – this calculation takes practice.

Things you need

Math graph paper or computer screen
Computer

Steps

Using Peak Finding Formula

Square’s complement

Advice

Warning

Things you need

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