Standard form to vertex form
In this minilesson, we will explore the process of converting standard form to vertex form and viceversa.
The standard form of a parabola is \(y=ax^2+bx+c\).
The vertex form of a parabola is \(y=a(xh)^2+k\).
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
Lesson Plan
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:
\(y=a(xh)^2+k\) 
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.
 In the vertex form, \((h,k)\) represents the vertex of the parabola where the parabola has either maximum/minimum value.
 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, \(y=a(xh)^2+k\), 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:
\[y=3 x^{2}6 x9\]
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 \(a(xh)^2+k\), which is in the vertex form.
Here, the vertex is, \((h,k)=(1,6)\).
If the above process seems difficult, then use the following steps:
 Compare the given equation with the standard form (\(y=ax^2+bx+c\)) and get the values of \(a,b,\) and \(c\).
 Apply the following formulas to find the values the values of \(h\) and \(k\) and substitute it in the vertex form (\(y=a(xh)^2+k\)):
\[ \begin{align} h&=\frac{b}{2 a}\\[0.2cm] k &= \frac{D}{4 a} \end{align}\]
Here, \(D\) is the discriminant where, \(D= b^24ac\).
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 stepbystep explanation.
How to Convert Vertex Form to Standard Form?
We know that the vertex form of parabola is \(y=a(xh)^2+k\).
To convert the vertex to standard form:

Example
Let us convert the equation \(y=3(x+1)^{2}6\) from vertex to standard form using the above steps:
\[\begin{align}
y&=3(x+1)^{2}6\\[0.2cm]
y&= 3(x+1)(x+1)6\\[0.2cm]
y&=3(x^2+2x+1)6\\[0.2cm]
y&=3x^26x36\\[0.2cm]
y&=3x^26x9\\[0.2cm]
\end{align} \]
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, \(a(xh)^2+k\): \[h=\dfrac{7}{4}\\[0.2cm] k=\dfrac{1}{8}\]
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: \[ D = b^24ac = 7^24(2)(6) = 1\]
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, \[y=a(x2)^2+2\,\,\,\rightarrow (1)\]
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):
\[ \begin{align} 0&=a(12)^2+2\\[0.2cm] 0&=a+2\\[0.2cm]a&=2 \end{align}\]
Substtute it back into (1) and expand the square to convert it into the standard form:
\[\begin{align}
y&=2(x2)^{2}+2\\[0.2cm]
y&= 2(x2)(x2)+2\\[0.2cm]
y&=2(x^24x+4)+2\\[0.2cm]
y&=2x^2+8x8+2\\[0.2cm]
y&=2x^2+8x6\\[0.2cm]
\end{align} \]
Thus, the standard form of the given parabola is:
\(y=2x^2+8x6\) 
Interactive Questions
Here are a few activities for you to practice.
Select/type your answer and click the "Check Answer" button to see the result.
Let's Summarize
The minilesson 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.
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Frequently Asked Questions (FAQs)
1. How to convert standard form to vertex form?
To convert standard form to vertex form, we just need to complete the square.
You can go to the "How to Convert Standard Form To Vertex Form?" section of this page to learn more about it.
2. How to convert vertex form to standard form?
To convert the vertex form to standard form:
 Expand the square, \((xh)^2\).
 Distribute \(a\).
 Combine the like terms.
You can go to the "How to Convert Vertex Form To Standard Form?" section of this page to learn more about it.
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 \(y=a(xh)^2+k\).
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.