The Vertex of a Parabola

Key IdeaThe graph of a quadratic function is a U-shaped curve called a parabola. One important feature of the graph is that it has an extreme point, called the vertex. If the parabola opens up, the vertex represents the lowest point on the graph, or the minimum value of the quadratic function. If the parabola opens down, the vertex represents the highest point on the graph, or the maximum value. In either case, the vertex is a turning point on the graph. The graph is also symmetric with a vertical line drawn through the vertex, called the axis of symmetry.

Vertex Formula

The standard form of the parabola is as follows.

y=ax2+bx+cy\,\,=\,\,ax^2\,\,+\,\,bx\,\,+\,\,c

 

The quadratic formula is a rule that says, in any equation of the form ax2 + bx + c = 0, the x-values of the equation are given by:

x=−b±b2−4ac2ax\,\,=\,\,\frac{-b\pm \sqrt{b^2-4ac}}{2a}

 

Where b2 – 4ac is called the discriminate.

We use the vertex formula to find the x– and y-coordinates of a parabola’s vertex. There are two ways to find the vertex of a parabola.

[1] The vertex can be calculated using the following formula.

(h,k)=(−b2a,−D4a)(h,k)\,=\,\left( \frac{-b}{2a},\,\frac{-D}{4a} \right)

 

Where “D” is the discriminant and is equal to D = b2 – 4ac, and “h” and “k” are the coordinates of the vertex. The above formula can also be written as follows:

(h,k)=(−b2a,c−b24a)\left( h, k \right) \,\,=\,\,\left( \frac{-b}{2a}, \,c\,-\,\frac{b^2}{4a} \right)

 

[2] The other method to find the vertex of a parabola is as follows:

  • We know that the x-coordinate of a vertex, i.e., h = -b/2a.
  • Now, substitute the x-coordinate value in the given standard form of the parabola equation y = ax2 + bx + c, we will get the y-coordinate (k) of a vertex.

Properties of Vertex of a Parabola

Here are some properties of the vertex of a parabola that follow from the definition of the vertex of a parabola.

  • The vertex of a parabola is its turning point.
  • Since the vertex of a parabola is its turning point, the derivative of the function representing the parabola at the vertex is 0.
  • A top/bottom open parabola either has a maximum or a minimum at its vertex.
  • The vertex of a left or right open parabola is neither a maximum nor a minimum to it.
  • Any type of parabola intersects its axis of symmetry at its vertex.

How Do You Derive “-b/2a” as the Equation for the x Value of the Vertex?

Completing the Square (Algebraic)

The standard form of the parabola is as follows.

y=ax2+bx+cy\,\,=\,\,ax^2\,\,+\,\,bx\,\,+\,\,c

 

Our goal is to rearrange this into vertex form: y = a(x – h)2 + k, where (h, k) is the vertex.

Factor out a⁠ from the first two terms:

y=a(x2+bax)+cy\,\,=\,\,a\left( x^2\,\,+\,\,\frac{b}{a}x \right) \,\,+\,\,c

 

Complete the square inside the parentheses. Take half of the x⁠-coefficient, b/2a⁠, and square to get

(b2a)2=b24a2\left( \frac{b}{2a} \right) ^2\,\,=\,\,\frac{b^2}{4a^2}

 

Add and subtract this inside the parentheses:

y=a(x2+bax+b24a2−b24a2)+cy\,\,=\,\,a\left( x^2\,\,+\,\,\frac{b}{a}x\,\,+\,\,\frac{b^2}{4a^2}\,\,-\,\,\frac{b^2}{4a^2} \right) \,\,+\,\,c

 

Move the subtracted term outside the parentheses. Remember to multiply it by the a on the outside:

y=a(x2+bax+b24a2)−b24a2+cy\,\,=\,\,a\left( x^2\,\,+\,\,\frac{b}{a}x\,\,+\,\,\frac{b^2}{4a^2} \right) \,\,-\,\,\frac{b^2}{4a^2}\,\,+\,\,c

 

Factor the perfect square trinomial inside the parentheses:

y=a(x+b2a)2+(c−b24a)y\,\,=\,\,a\left( x\,\,+\,\,\frac{b}{2a} \right) ^2\,\,+\,\,\left( c\,\,-\,\,\frac{b^2}{4a} \right)

 

Comparing this to the vertex form y = a(x – h)2 + k, we can see that:

−h=b2a⇒h=−b2a-h\,\,=\,\,\frac{b}{2a}\,\,\Rightarrow \,\,h\,\,=\,\,-\frac{b}{2a}

 

Since h represents the x⁠-coordinate of the vertex, x = -(b/2a).

Using Calculus (Optimization)

The vertex of a parabola is its highest (maximum) or lowest (minimum) point. At this exact turning point, the tangent line is perfectly horizontal, meaning the slope (derivative) is zero.

Take the derivative of the standard equation y = ax2 + bx + c with respect to x⁠:

dydx=2ax+b\frac{dy}{dx}\,\,=\,\,2ax\,\,+\,\,b

 

Set the derivative to zero to find the turning point:

2ax+b=02ax\,\,+\,\,b\,\,=\,\,0

 

Solve for x:

2ax=−b2ax\,\,=\,\,-b

 

x=−b2ax\,\,=\,\,-\frac{b}{2a}

QED

References

Brummer, Josh, and Taran Funk, et al. Mathematics. PreCalculus Mathematics at Nebraska. University of Nebraska, October 29, 2026. https://mathbooks.unl.edu/PreCalculus/Section-The-Vertex-of-a-Parabola.html.

“Characteristics of Parabolas.” College Algebra. lumen. Accessed November 7, 2026. https://courses.lumenlearning.com/waymakercollegealgebra/chapter/characteristics-of-parabolas/.

Stapel, Elizabeth. “The Quadratic Formula Explained.” Purplemath. Accessed November 7, 2026. https://www.purplemath.com/modules/quadform.htm.

“Vertex (Parabola).” Math Is Fun. Accessed November 7, 2026. https://www.mathsisfun.com/definitions/vertex-parabola-.html.

“Vertex Formula.” BYJU’S, September 17, 2020. https://byjus.com/vertex-formula/.

“Vertex Formula – What Is Vertex Formula? Examples.” CUEMATH. Accessed November 7, 2026. https://www.cuemath.com/vertex-formula/.

[ ℰ ] “Vertex of a Parabola – Formula.” CUEMATH. Accessed November 7, 2026. https://www.cuemath.com/geometry/vertex-of-a-parabola/.

[ ℰ ] This exceptional reference is highly recommended for your consideration.

The feature image on this page is from the CUEMATH website.

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