Standard form to vertex form

Standard form to vertex form
Go back to  'Standard Form'

In this mini-lesson, we will explore the process of converting standard form to vertex form and vice-versa.

The standard form of a parabola is $$y=ax^2+bx+c$$.

The vertex form of a parabola is $$y=a(x-h)^2+k$$.

Here, the vertex form has a square in it.

So to convert the standard to vertex form we need to complete the square.

Let's learn about the following in detail:

• Standard form to vertex form
• Vertex form to standard form

How to Convert Standard Form To Vertex Form?

Standard Form

The standard form of a parabola is:

 $$y=ax^2+bx+c$$

Here, $$a, b,$$ and $$c$$ are real numbers (constants) where $$a \neq 0$$.

$$x$$ and $$y$$ are variables where $$(x,y)$$ represents a point on the parabola.

Vertex Form

The vertex form of a parabola is:

 $$y=a(x-h)^2+k$$

Here, $$a,h,$$ and $$k$$ are real numbers where $$a\neq 0$$.

$$x$$ and $$y$$ are variables where $$(x,y)$$ represents a point on the parabola.

Important Notes
1. In the vertex form, $$(h,k)$$ represents the vertex of the parabola where the parabola has either maximum/minimum value.
2. If $$a>0$$, the parabola has minimum value at $$(h,k)$$ and
if $$a<0$$, the parabola has maximum value at $$(h,k)$$.

Standard to Vertex Form

In the vertex form, $$y=a(x-h)^2+k$$, there is a "whole square."

So to convert the standard form to vertex form, we just need to complete the square.

Let us learn how to complete the square using an example.

Example

Convert the parabola from standard to vertex form:

$y=-3 x^{2}-6 x-9$

Solution:

First, we should make sure that the coefficient of $$x^2$$ is $$1$$

If the coefficient of $$x^2$$ is NOT $$1$$, we will place the number outside as a common factor.

We will get:

$y=-3 x^{2}-6 x- 9 = -3 \left(x^2+2x+3\right)$

Now, the coefficient of $$x^2$$ is $$1$$

Step 1: Identify the coefficient of $$x$$.

Step 2: Make it half and square the resultant number.

Step 3: Add and subtract the above number after the $$x$$ term in the expression.

Step 4: Factorize the perfect square trinomial formed by the first 3 terms using the suitable identity

Here, we can use $$x^2+2xy+y^2=(x+y)^2$$.

In this case, $x^2+2x+ 1= (x+1)^2$

The above expression from Step 3 becomes:

Step 5: Simplify the last two numbers and distribute the outside number.

Here, $$-1+3=2$$

Thus, the above expression becomes:

This is of the form $$a(x-h)^2+k$$, which is in the vertex form.

Here, the vertex is, $$(h,k)=(-1,-6)$$.

Tips and Tricks

If the above process seems difficult, then use the following steps:

1. Compare the given equation with the standard form ($$y=ax^2+bx+c$$) and get the values of $$a,b,$$ and $$c$$.
2. Apply the following formulas to find the values the values of $$h$$ and $$k$$ and substitute it in the vertex form ($$y=a(x-h)^2+k$$):
\begin{align} h&=-\frac{b}{2 a}\\[0.2cm] k &= -\frac{D}{4 a} \end{align}
Here, $$D$$ is the discriminant where, $$D= b^2-4ac$$.

Standard Form to Vertex Form Calculator

Here is the "Standard Form to Vertex Form Calculator."

You can enter the equation of the parabola in the standard form. This calculator shows you how to convert it into the vertex form with a step-by-step explanation.

How to Convert Vertex Form to Standard Form?

We know that the vertex form of parabola is $$y=a(x-h)^2+k$$.

To convert the vertex to standard form:

 Expand the square, $$(x-h)^2$$. Distribute $$a$$. Combine the like terms.

Example

Let us convert the equation $$y=-3(x+1)^{2}-6$$ from vertex to standard form using the above steps:
\begin{align} y&=-3(x+1)^{2}-6\\[0.2cm] y&= -3(x+1)(x+1)-6\\[0.2cm] y&=-3(x^2+2x+1)-6\\[0.2cm] y&=-3x^2-6x-3-6\\[0.2cm] y&=-3x^2-6x-9\\[0.2cm] \end{align}

Solved Examples

 Example 1

Can we help Sophia to find the vertex of the parabola $$y=2 x^{2}+7 x+6$$ by completing the square?

Solution

The given equation of parabola is $$y=2 x^{2}+7 x+6$$.

To complete the square, first, we will make the coefficient of $$x^2$$ as $$1$$

We will take the coefficient of $$x^2$$ (which is $$2$$) as a common factor.

$2 x^{2}+7 x+6 = 2\left( x^2 + \dfrac{7}{2}x+ 3 \right) \,\,\,\,\,\rightarrow (1)$

The coefficient of $$x$$ is $$\dfrac{7}{2}$$

Half of it is $$\dfrac{7}{4}$$

Its square is $$\left( \dfrac{7}{4} \right)^2= \dfrac{49}{16}$$

This term can also be found using $$\left( \dfrac{-b}{2a}\right)^2 = \left( \dfrac{-7}{2(2)} \right)^2= \dfrac{49}{16}$$

Add and subtract it after the $$x$$ term in (1):

$2 x^{2}\!+\!7 x\!+\!6 = 2\left(\!\!x^2 \!+\! \dfrac{7}{2}x\!+\!\dfrac{49}{4}\!-\!\dfrac{49}{4} +3 \!\!\right)$

Factorize the trinomial made by the first three terms:

\begin{aligned}&2 x^{2}\!+\!7 x\!+\!3\!\\[0.2cm] &= 2\left( \!x^2 + \dfrac{7}{2}x+\dfrac{49}{16}-\dfrac{49}{16}+3\! \right)\\[0.2cm] &= 2 \left(\!\! \left(x+ \dfrac{7}{4} \right)^2 -\dfrac{49}{16}+3 \right)\\ &= 2 \left( \left(x+ \dfrac{7}{4} \right)^2 -\dfrac{1}{16} \right)\\ &= 2\left(x+ \dfrac{7}{4} \right)^2 - \dfrac{1}{8} \end{aligned}

By comparing the final equation with the vertex form, $$a(x-h)^2+k$$: $h=-\dfrac{7}{4}\\[0.2cm] k=-\dfrac{1}{8}$

Thus the vertex of the given parabola is:

 $$(h,k)= \left(-\dfrac{7}{4},-\dfrac{1}{8}\right)$$
 Example 2

Though we helped Sophia to find the vertex of $$y=2 x^{2}+7 x+6$$ in the above example, she is still not comfortable with this method.

Can we help her to find its vertex without completing the square?

Solution

The given equation of parabola is $$y=2 x^{2}+7 x+6$$.

We will use the trick mentioned in the Tips and Tricks section of this page to find the vertex without completing the square.

Compare the given equation with $$y=2 x^{2}+7 x+6$$:

\begin{align} a&=2\\[0.2cm]b&=7\\[0.2cm]c&=6 \end{align}

The discriminant is: $D = b^2-4ac = 7^2-4(2)(6) = 1$

We will find the coordinates of the vertex using the formulas:

\begin{align} h&=-\frac{b}{2 a}=- \dfrac{7}{2(2)} =- \dfrac 7 4\\[0.2cm] k &= -\frac{D}{4 a}= -\dfrac{1}{4(2)}= - \dfrac{1}{8} \end{align}

Therefore, the vertex of the given parabola is:

 $$(h,k)= \left(-\dfrac{7}{4},-\dfrac{1}{8}\right)$$

Note that the answer is same as that of Example 1.

 Example 3

Find the equation of the following parabola in the standard form:

Solution

We can see that the parabola has the maximum value at the point $$(2,2)$$.

So the vertex of the parabola is, $(h,k)=(2,2)$

So the vertex form of the above parabola is, $y=a(x-2)^2+2\,\,\,\rightarrow (1)$

To find $$a$$ here, we have to substitute any known point of the parabola in this equation.

The graph clearly passes through the point $$(x,y)=(1,0)$$.

Substitute it in (1):

\begin{align} 0&=a(1-2)^2+2\\[0.2cm] 0&=a+2\\[0.2cm]a&=-2 \end{align}

Substtute it back into (1) and expand the square to convert it into the standard form:

\begin{align} y&=-2(x-2)^{2}+2\\[0.2cm] y&= -2(x-2)(x-2)+2\\[0.2cm] y&=-2(x^2-4x+4)+2\\[0.2cm] y&=-2x^2+8x-8+2\\[0.2cm] y&=-2x^2+8x-6\\[0.2cm] \end{align}

Thus, the standard form of the given parabola is:

 $$y=-2x^2+8x-6$$

Interactive Questions

Here are a few activities for you to practice.

Let's Summarize

The mini-lesson targeted the fascinating concept of Standard Form to Vertex Form. The math journey around Standard Form to Vertex Form starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever. Here lies the magic with Cuemath.

At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students!

Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic.

Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in.

1. How to convert standard form to vertex form?

To convert standard form to vertex form, we just need to complete the square.

2. How to convert vertex form to standard form?

To convert the vertex form to standard form:

• Expand the square, $$(x-h)^2$$.
• Distribute $$a$$.
• Combine the like terms.

3. How to find the vertex of a parabola in standard form?

To find the vertex of a parabola in standard form, first, convert it to the vertex form $$y=a(x-h)^2+k$$.

Then $$(h,k)$$ would give the vertex of the parabola.

Example 1 and Example 2 under the "Solved Examples" section of this page is related to this. Check this out.

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