Find the Vertex by Using the Quadratic Formula

The Quadratic Formula is primarily used to identify the roots (\(x\)-intercepts) of a quadratic function. What many people don’t know is that you can also easily find the vertex of the function by simply looking at the Quadratic Formula!


Graphing Quadratic Functions

When graphing quadratic functions by hand, we will often use the quadratic formula to obtain the roots (or \(x\)-intercepts) of the function. Then separately, we will work on finding the vertex of the function since this point defines the lowest (or highest) point on the graph, as well as the line of symmetry.

Graph of a generic quadratic function showing both roots and the vertex.

For the general quadratic function, \(f(x) = ax^2 + bx + c\), the quadratic formula identifies the roots at:

x = \frac{-b \pm\sqrt{b^2 – 4ac}}{2a}
while the the vertex is found at \(\left(-\frac b {2a}, f\left(-\frac b {2a}\right)\right)\).

The \(x\)-value of the vertex, \(x = -\frac b {2a}\), is often referred to in algebra textbooks as The Vertex Formula.


Finding the Vertex in the Quadratic Formula

Many people notice (or are taught) that the Vertex Formula can be found nestled in the Quadratic Formula. Specifically,

\mbox{Quadratic Formula: } x = \frac{\color{blue}{-b} \pm\sqrt{b^2 – 4ac}}{\color{blue}{2a}}\longrightarrow \color{blue}{\frac{-b}{2a}} = x\mbox{-value of the vertex.}

Less well known is the fact that the \(y\)-coordinate of the vertex can also be obtained easily from the quadratic formula. The vertex of a quadratic function can be written as

\mbox{Vertex: } \left(\color{blue}{-\frac b {2a}}, \color{red}{-\frac D {4a}}\right)
where \(D\) is the discriminant of the quadratic function. Recall that the discriminant of a quadratic function is \(\color{red}{D = b^2 – 4ac}\) and it appears under the radical in the quadratic formula:

\mbox{Quadratic Formula: } x = \frac{-b \pm\sqrt{\color{red}{b^2 – 4ac}}}{2a} = \frac{-b \pm\sqrt{\color{red} D}}{2a}.

Also, notice that the denominator of the \(x\)-coordinate and \(y\)-coordinate of the vertex are very similar.  The \(y\)-coordinate’s denominator is just twice as big as the \(x\)-coordinate’s!

Now, let’s try out this idea and find the vertex of a quadratic from just the quadratic formula.

Example 1:

Suppose \(f(x) = 5x^2 + 4x – 3\). Use the quadratic formula to identify the entire vertex of the function, and then verify the answer using the standard method.

Solution: The quadratic formula for this function is
x = \frac{-4 \pm \sqrt{4^2 – 4(5)(-3)}}{2(5)} = \frac{-4\pm \sqrt{76}}{10}.

The discriminant for this function is \(\color{red}{D = 76}\). So, the vertex should be at

\mbox{Vertex: } \left(\color{blue}{-\frac b {2a}}, \color{red}{-\frac D {4a}}\right) = \left(-\frac 4 {10}, -\frac{\color{red}{76}}{20}\right) = \left(-\frac 2 5, -\frac{19} 5\right).

Let’s double check our \(y\)-coordinate by using the standard method (plugging \(x = -\frac b {2a} = -\frac 2 5\) into the original function).

f\left(\color{blue}{-\frac 2 5}\right)
& = 5\left(\color{blue}{-\frac 2 5}\right)^2 + 4\left(\color{blue}{-\frac 2 5}\right) – 3 && \mbox{Plugging in the \(x\)-value.}\\[6pt]
& = 5\left(\frac 4 {25}\right) – \frac 8 5 – \frac 3 1 && \mbox{Simplifying.}\\[6pt]
& = \frac 4 5 – \frac 8 5 – \frac{15} 5\\[6pt]
& = \frac{-19} 5 && \mbox{\(y\)-value of the vertex.}


Example 2

Let’s try another example. Suppose \(f(x) = -2x^2 +12x -25\). Identify the vertex using the quadratic formula, and verify the answer with the standard method.

Solution: The quadratic formula for this function is

x = \frac{-12 \pm\sqrt{(-12)^2-4(-2)(-25)}}{2(-2)} = \frac{\color{blue}{-12} \pm\sqrt{\color{red}{-56}}}{-4}

The vertex should be \(\displaystyle \left(\frac{\color{blue}{-12}}{-4}, -\frac{\color{red}{-56}}{-8}\right) = \left(3, -7\right)\).  Let’s verify the result. Using the standard method, we evaluate \(f(3)\).

f(\color{blue} 3) & = -2(\color{blue} 3)^2 + 12(\color{blue} 3) – 25\\[6pt]
& = -2(9) + 36 – 25\\
& = -18 + 11\\
& = -7

Again, we see the method works.

However, the savvy math student will immediately ask, “Does it work every time? Or was there something `special’ or `unusual’ about these particular quadratic functions?”


Does it Always Work?

This is an important question. In order to claim something is valid in mathematics, we have to do more than look at a few examples. We have to prove its validity. So, without further ado, here is the proof of this technique.

Given: The \(x\)-coordinate of the vertex of a quadratic function is \(x = -\frac b {2a}\).

Show: The \(y\)-coordinate of the vertex of a quadratic function is \(y = -\frac D {2a}\), where \(D\) is the discriminant of the function.

Proof: Simply evaluate \(f(x) = ax^2 + bx + c\) at \(x = -\frac b {2a}\).

f\left(\color{blue}{-\frac b {2a}}\right)
& = a\left(\color{blue}{-\frac b {2a}}\right)^2 + b\left(\color{blue}{-\frac b {2a}}\right) + c\\[6pt]
& = a\left(\frac{b^2}{4a^2}\right) -\frac{b^2}{2a} + c && \mbox{Simplify a little.}\\[6pt]
& = \frac{b^2}{4a} -\frac{b^2}{2a} + \frac c 1 && \mbox{Cancel the common factor.}\\[6pt]
& = \frac{b^2}{4a} -\frac{2b^2}{4a} + \frac{4ac}{4a} && \mbox{Common denominator.}\\[6pt]
& = \frac{-b^2 + 4ac}{4a} && \mbox{Combine numerators.}\\[6pt]
& = -\frac{\color{red}{b^2 – 4ac}}{4a} && \mbox{Factor out a negative 1.}\\[6pt]
& = -\frac{\color{red}{D}}{4a}

Well, how about that. It seems like any time we evaluate a quadratic at \(x = -\frac b {2a}\), we will end up with \(-\frac D {4a}\).


So What?

So, what does this do for us in practical terms? Suppose we were asked to graph (by hand) the function \(f(x) = 3x^2 -6x + 8\). A good first step is to find the roots, and a good method for that is to use the quadratic formula.

x & = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\\[6pt]
& = \frac{6 \pm \sqrt{(-6)^2 – 4(3)(8)}}{2(3)}\\[6pt]
& = \frac{6 \pm \sqrt{-60}} 6

Oops. This quadratic function doesn’t have real-valued \(x\)-intercepts (Yes, it has complex roots, but that won’t help us graph, will it?).

But the function has a graph, we just have to find it. If only we had the vertex and one other point! Then we could graph the quadratic! But wait! We already have written out the quadratic formula, so that should tell us the vertex:

\mbox{Vertex: } \left(-\frac b {2a}, -\frac D {4a}\right) = \left(-\frac 6 6, -\frac{-60}{12}\right) = (1, 5).

Now, together with the \(y\)-intercept at \((0,8)\), we can easily plot the function.

Graph of the quadratic function. The vertex was easily found using only the quadratic formula.



In conclusion, the more we understand about the tools we have at our disposal, the easier it becomes to work through the mathematics that crop up in school and careers. And the more we learn, the more we become intrigued by the idea of other possibilities.  I mean, doesn’t this insight into the quadratic formula just beg the question, “What else can we learn from the Quadratic Formula? Or the discriminant?”


© HT Goodwill and, 2016. Unauthorized use and/or duplication of this material without express and written permission from this site’s author and/or owner is strictly prohibited. Excerpts and links may be used, provided that full and clear credit is given to HT Goodwill and with appropriate and specific direction to the original content.

10 Ways Tutoring Can Help You

10 Ways Tutoring Can Help You

Finding a good math tutor can be essential to your success in school. Tutoring can do much more than just improving math grades. Here are 10 ways in which you can benefit from tutoring.

1. Improve your confidence

When working one-on-one outside of the classroom setting, you might find it easier to focus on the subject material and return to math class with increased confidence.

2. Improve your grades 

Supplementing your coursework with tutoring can help you understand the material more thoroughly so you can do better in school.

3. Work at a pace you are comfortable with 

Your teacher might be moving through the material at a speed that is too fast for you, but when you are studying with a tutor they will structure their sessions at a pace that makes you feel comfortable. Alternatively, if you feel that your class is moving too slow, your tutor can work ahead with you to prepare for future material.

4. Build mathematical foundation

It is essential that you understand the basics before you can succeed in more difficult math. While your teacher must follow a strict curriculum,a tutor will work with you to review the material that you didn’t understand before. Building a strong mathematical foundation now will allow you to tackle the more complicated topics later.

5. Provide supplemental work

If you are confused about a certain topic, your tutor can find you extra material and work through it with you.

6. Tailor lessons to your needs 

In a classroom setting, students work at different paces and understand things differently. In one-on-one tutoring, your lessons can be perfectly tailored to your own learning style, which can make learning math easier and more enjoyable.

7. Find a new challenge 

Is your math class too easy for you? Tutoring might be the perfect way to push your own boundaries. A tutor can expand and enrich upon the material you’re learning in school, or work on interesting material that’s not even offered at your school. Your tutor can also help you find local math competitions to challenge you in a fun way.

8. Improve SAT/ACT scores 

Tutoring is a great way to make sure you ace your college entrance exam. Tutors can help you learn test taking strategies, as well as helping you learn what kind of questions to expect.

9. Bring a fresh new perspective 

If you’re feeling stuck in a mathematical rut, working with a tutor can help. Your tutor can discuss math from a new point of view that might make more sense to you than the way your teacher approaches the material.

10. Make math more interesting 

Tutors are passionate about math and would love to share that passion with you. If there is a certain topic of interest to you, your tutor can build some lessons around that to keep you interested.