## Vertex Form of a Quadratic Equation

The vertex form of a quadratic equation is a wonderful thing! As you’ll soon find out, finding the coordinate of the vertex from standard form is quite difficult. But, here with vertex form, we can just find it by inspection, similar to finding the *y* – intercept from standard form.

In addition, finding the *x* – intercepts from standard form does not require factoring, one can just use inverse operations to solve for *x*. To find the *y* – intercept requires more work compared to finding it from standard form, but not much. It is a calculation that can be done *in your head*, most of the time.

Read through the notes below, watch the video, try the practice problems and be good to go! As always, please leave questions, comments and the like in the comments in the video. I’d love to hear from you!

There are two ways of writing a quadratic equation that are particularly useful. The first is called Standard Form, it is *ax*^{2} + *bx + c*, where *a, b, *and *c* are coefficients. In the previous section we saw that from Standard Form one can find the *x* – intercepts by setting *y* = 0 and factoring to solve. The *y* – intercept is (0, *c*), and the vertex can be found with the formula below.

Standard form is easy for finding the *y* – intercept and if factorable, can be easy to find the *x* – intercepts. Further, there are other uses that are offered from Standard Form that we’ll explore later.

Vertex Form is not as commonly seen, but it is very useful. The easiest thing about Vertex Form is how easy it is to find the vertex, which is the most difficult thing to do from Standard Form. Let’s see what Vertex Form is, first, then talk about how to find the vertex, then the *x* – intercepts, and last, the *y* – intercept.

Vertex Form: *y* = *a*(*x – h*)^{2} + *k*

Notice the only coefficient named the same as is done with Standard Form is the leading coefficient, *a*. If a specific equation is written in Standard or Vertex Form, the leading coefficient will be exactly the same. The other numbers will be different.

Let’s see an example. The following equation is in Vertex Form.

To change this into Standard form we would just need to follow the order of operations. That is, square the (*x* – 1)¸then distribute the 2, then add 3.

So, the following two are exactly the same value, just written in a different form.

From Standard Form, the *y* – intercept can be read, it is (0, *c*). So our *y* – intercept here is (0, 5). You cannot just read the *y* – intercept’s value from Vertex Form, you have to calculate it. But, you can read the vertex. The vertex is (*h*, *k*).

Let’s look at our equation again: *f*(*x*) = 2(*x* – 1)^{2} + 3. Here *a* = 2, *h* = 1, and *k* = 3. The confusing part is the value of *h*. The equation has – 1, which makes *h* equal to + 1. In the generic form of Vertex Form, *y = a*(*x* – *h*)^{2} + *k*, you can see the binomial being squared is *x – h*. So, whatever number you see there, you have to take the opposite sign as the value of *h*.

Since the vertex is (*h¸ k*), and *h *= 1, and *k* = 3, our vertex is (1, 3). You can use the vertex formula to find the vertex of the equation *y* = 2*x*^{2} – 4*x* + 5. It is the same vertex because the equation is the same value, just written in a different form.

To find the *y* – intercept we must substitute zero for *x*, and perform the calculation. This is exactly what we do when finding *y* – intercepts from Standard Form, except the calculation is so obvious we just forget about it. Let’s see how it looks to find the *y* – intercept from Vertex Form. (The intercept should be (0, 5), as can be seen from Standard Form.)

Here we just need to follow the order of operations, and we’ll be good to go.

The vertex is much easier to find from Vertex Form, compared to Standard Form. The *y* – intercept is slightly trickier, but not difficult. Now, let’s discuss finding the *x* – intercepts from Vertex Form.

From Standard Form you will learn three ways to find *x* – intercepts. You’ve already seen factoring, which can be difficult and does NOT always work. There can be *x* – intercepts for equations that are NOT factorable.

If the equation has *x* – intercepts, they can be found from Vertex Form, 100% of the time. To find the *x *– intercepts, we set the equation equal to zero ( *f*(*x*) = 0), and then solve for *x*. Here we can use inverse operations. Let’s see an example.

Using inverse operations will give us the same steps every time. We’ll subtract (or add) *k* from each side, then divide by *a*. Then, we’ll take the square root of both sides, and then add (or subtract) *h*. Let’s see it in action.

This number is Imaginary. The square root of positive numbers are all Real Numbers and will yield *x* – intercepts (in this application). If, when finding the *x* – intercepts from Vertex Form, you find a negative square root, check for sign errors. If none were made, then the equation does NOT have *x* – intercepts.

It makes sense that this equation does NOT have *x* – intercepts. The vertex is (1, 3), which is above the *x* – axis. Since *a* is positive 2, the graph will go up, making the vertex the lowest point on the parabola. The lowest point is above the *x* – axis, and it goes up, so it will not cross the *x* – axis. Therefore, no *x* – intercepts!

Let’s see an example that does have *x* – intercepts.

When you divide, make sure you reduce. But, do NOT change to a decimal or write as a mixed number. Math is a lot easier to perform with reduced improper fractions!

Here we have two values for the square root of 9/4s. Remember, the square root of a number asks, “*What squared equals the radicand?*”. Well, for 9, there are two numbers squared that equal 9. First, 3 × 3 = 9, but also -3 × -3 = 9. For 4, there are two numbers, 2 and -2. So, for 9/4, we have a + 3/2 and a – 3/2. That’s what the symbol “plus over minus,” refers two, there being two answers.

To solve this, we need to write each equation out.

A quick and easy way to think about these fractions is to think of what fraction with a denominator of two equals 2…which is of course 4/2. 3/2 – 4/2 = -1/2, and -3/2 – 4/2 = – 7/2.

Our *x* – intercepts are:

Summary: Vertex Form can be changed to Standard Form by just following the order of operations. The vertex of a quadratic equation can be read from Vertex Form, it is (*h, k*). The only trick there is to remember that *h* is the opposite sign of what is written. So, for *y* = 3(*x* + 1)^{2} – 2, the vertex is (-1, -2).

The *y* – intercept is still found by replacing *x* with zero and simplifying. The *x* – intercepts can be found by using inverse operations.