Vertex form of a parabola,Standard to vertex form conversion

Contents:

This article covers the vertex form of a parabola, how to find the vertex of a parabola and the intercepts.

Please do attempt the QUIZ at the end of the article to ensure complete understanding of the topic.

The following is the brief of key concepts and learning objectives :

  • How to find the vertex of a parabola and its coordinates?
  • How to find the X-intercepts and Y-intercepts?
  • What is the vertex form of a parabola?
  • How to convert the standard form of a parabola to the vertex form?
  • How to find the Maximum /Minimum of the parabola using the vertex?
  • Worked out examples.
  • Quiz (with worked out solutions).
  • “Free math Help Forum”- to get answers to your questions from experts.

1)How to find the vertex of a parabola?

Graph of the quadratic function y= f\left( x\right) =ax^{2}+bx+c is called as a parabola.

The Vertex of a parabola can easily be identified, when the graph of the same is given. It is a point, where there would be either minimum or maximum value of the Parabola.

But how do we find the coordinates of the vertex from a given quadratic function?.

Let’s say (h,k) are the coordinates of the vertex of a parabola, where “h” is the “x” coordinate and “k” is the “y” co-ordinate coordinate.

The “h” can be found using the formula  h=-\dfrac {b}{2a}.where we can get the coefficients a,b,c on comparion with the standard form of the quadratic .

Once we get the value of “h”, we can plug it into the quadratic function and get the value of “k”.

Plugging in x=-\dfrac {b}{2a},y=k

\begin{aligned}y=ax^{2}+bx+c\\\Rightarrow k=a\left( -\dfrac {b}{2a}\right) ^{2}+b\left( -\dfrac {b}{2a}\right) +c\\ \Rightarrow k=a\left( \dfrac {b^{2}}{4a^{2}}\right) -\dfrac {b^{2}}{2a}+c\\ \Rightarrow k=\dfrac {b^{2}}{4a}-\dfrac {b^{2}}{2a}+c\\ \Rightarrow k =-\dfrac {b^{2}}{4a}+c\end{aligned}

Here is a quick example for easy understanding.Let’s say we have  f\left( x\right) =2x^{2}+8x+6.

On comparing with the standard form, a= 2 and b=8 c= 6.Therefore x -coordinate of the vertex h = -\dfrac {b}{2a}=-\dfrac {8}{2\times 2}=-4

Since we have x-coordinate h= -4, we plug x =-4 into the function  f\left( x\right) =2x^{2}+8x+6.

\begin{aligned}f\left( -4\right) =2\left( -4\right) ^{2}+8\left( -4\right) +6\\ f\left( -4\right) =32-32+6\\ f\left( -4\right) =6\end{aligned}.

Therefore the vertex of given  f\left( x\right) =2x^{2}+8x+6 is (h,k)=(-4,6)

2) How to find the X-intercepts and Y-intercepts of a parabola?

 X-Intercepts:-

  • These are the points where the parabola, intersects the  “x” axis. Since it is a point of “x’- axis, the “y” coordinate would be equal to 0.There can be at most two X-intercepts for a quadratic function.
  • Finding the “x”-intercepts is nothing but finding the roots of a quadratic function.
  • We can use either Factorization method or Quadratic formula to find the roots of a quadratic function, which are explained in detail.

Y-Intercepts:-

  • Finding the “y”-Intercepts is quite simpler when compared to that of finding “x”- intercepts. Since it is a point on the y-axis, we plug in x =0 and solve for the “y”.

Let us consider an example of  f\left( x\right) =x^{2}+3x+4.

X-intercept:

we plug in f(x) =0, therefore making  x^{2}+3x+4 =0

Using  Factorization method as discussed in the previous article,we would get (x+3)(x+1) =0.Therefore solving for x we have x =-3 and x =-1.

Hence we can conclude that the “x”- intercepts are (-3,0) and ( -1,0).

 

Y-Intercept:

We plug in x=0 in  f\left( x\right) =x^{2}+3x+4.

Therefore f(0)=  0^{2}+0*x+4 =4

Hence we can conclude that the Y-intercept for the given quadratic is (0,4).

 

3)Vertex form of a parabola.

Vertex form of a parabola gives the equation of the parabola in terms of vertex coordinates(h,k). If we identify the vertex of a quadratic, we can just plug it in the formula and get the equation.

Vertex form of a parabola is given by y=a\left( x-h\right) ^{2}+k or x=a\left( y-k\right) ^{2}+h depending upon the orientation of the parabola.

4)The standard form to vertex form of a parabola conversion.

We learned the formula to find the vertex “x”-coordinate “h” of quadratic where    h=-\dfrac {b}{2a}.

The question is how did we arrive at that formula? How did the vertex form of parabola evolve?

Vertex form is just another form of quadratic function  y=ax^{2}+bx+c obtained by completing the square.

These steps are very similar to that of  Solving the quadratic equation by completing the square method.

Let us look step by step how we can get the Vertex form of a parabola from the Standard form  y=ax^{2}+bx+c

Step 1: Ensure that the coefficient of   x^{2} is 1.If not divide the equation throughout with the coefficient.

  • Since we have the coefficient as “a”  let us divide the equation throughout by “a”.

\begin{aligned}\dfrac {y}{a}=\dfrac {ax^{2}}{a}+\dfrac {b}{a}x+\dfrac {c}{a}\\ \Rightarrow \dfrac {y}{a}=x^{2}+\dfrac {b}{a}x+\dfrac {c}{a}\end{aligned}

 

Step 2: Find the coefficient of “x” and divide it by 2.

  • The coefficient of “x” is \dfrac {b}{a}.Therefore dividing by 2 we get \dfrac {b}{2a}.

 

Step 3:Add and subtract  \left( \dfrac {b}{2a}\right) ^{2} to the quadratic equation to complete the square.

  • Therefore , the quadratic equation would be

\begin{aligned}\dfrac {y}{a}=x^{2}+\dfrac {b}{a}x+\left( \dfrac {b}{2a}\right) ^{2}-\left( \dfrac {b}{2a}\right) ^{2}+\dfrac {c}{a}\\ \Rightarrow \dfrac {y}{a}=\left( x+\dfrac {b}{2a}\right) ^{2}-\dfrac {b^{2}}{4a^{2}}+\dfrac {c}{a}\end{aligned}\\\begin{aligned}\Rightarrow y=a\left( x-\left( -\dfrac {b}{2a}\right) \right) ^{2}-a\left( \dfrac {b^{2}}{4a^{2}}\right) +a\dfrac {c}{a}\\\Rightarrow y=a\left( x-\left( -\dfrac {b}{2a}\right) \right) ^{2}-\dfrac {b^{2}}{4a}+c\end{aligned}.

Step4: Replace the vertex coordinates h and k to get the vertex form.

  • On comparision with the  vertex form of the parabola y=a\left( x-h\right) ^{2}+k ,h=-\dfrac {b}{2a},k =-\dfrac {b^{2}}{4a}+c.

 

5) How to find the Maximum /Minimum of the parabola using the vertex?

As discussed in the previous article, the vertex is either maximum or minimum of the parabola.

The “y” coordinate of the vertex, gives the Maximum or Minimum value of a parabola always.

Therefore, even without graphing the quadratic function, we can find the extremes of a quadratic by plugging the coefficients a,b,c into the formula k =-\dfrac {b^{2}}{4a}+c.

6)Summary:

  • Co-ordinates of Vertex of a parabola can be found form the coefficients a,b,c of the standard form of quadratic y=ax^{2}+bx+c .
  • The “x”-Coordinate of vertex of a parabola is h=-\dfrac {b}{2a}and “y”-Coordinate is k =-\dfrac {b^{2}}{4a}+c.
  • The “y”-Coordinate “K” always gives the information of the maximum or minimum of the parabola.
  • If we have the standard form of the parabola, we can find the vertex coordinates h,k and write the Vertex form of a parabola, which is given by y=a\left( x-h\right) ^{2}+k.

 

7)Worked out Examples:

Example-1 

Find the vertex form from the standard form of the parabolay=2x^{2}+8x+6

Strategy Step-by-Step:-

Step1: Compare the given problem with the standard form and get the coefficients a,b,c.

  • On comparison the standard form y=ax^{2}+bx+c a=2,b=8 c=6.

 

Step2: Ensure the coefficient a=1.If not pull out the common factor “a” out of all the terms.

  • Since the a=2 we need to take 2 as the common factor.

\begin{aligned}y=2x^{2}+8x+6\\ y=2\left( x^{2}+4x+3\right) \end{aligned}

 

Step3:Get the new coefficient “b”.Add and subtract  \left( \dfrac {b}{2}\right) ^{2}

  • The new value of b= 4 therefore  \left( \dfrac {4}{2}\right) ^{2}=2^{2} =4.

 

Step4: Complete the square and move the unwanted numbers out of the parenthesis to get the vertex form.

  • we have the unwanted terms -4+3 inside the parenthesis .So -4+3=-1and we move it out which gives -2.

\begin{aligned}y=2\left( x^{2}+4x+4\left( -4+3\right) \right) \\ y=2\left( x^{2}+4x+4\right) -2\\ y=2\left( x+2\right) ^{2}-2\end{aligned}.

Therefore the vertex form of the quadratic is y=2\left( x+2\right) ^{2}-2\end{aligned}.Hence on comparision the vertex coordinates would be (-2,-2).

Example-2:

i)Find the vertex of the parabola y=2x^{2}+8x+6  using formulas and write the vertex form. ii)Find xy-intercepts.

It is quite easy to write the vertex form of a quadratic using the formulas, instead of deriving it from completing the square method. All we need is to find the vertex coordinates h and k and plug it into the formula.

i)Finding the vertex of the parabola using formulas

Strategy Step-by-Step:-

Step1: Compare the given problem with the standard form and get the coefficients a,b,c.

  • Therefore,On comparision, with the standard form y=ax^{2}+bx+c a=2,b=8 c=6.

 

Step2: To get the “x”- intercept use the formula -\dfrac {b}{2a}.

  • Plugging in the value of “b” as 8  and “a” as 2 into the formula we get

-\dfrac {b}{2a}=\dfrac {8}{2\left( 2\right) }=-2.

Therefore the x-coordinate of the vertex is -2

 

Step3: Plug in the value of the “x-coordinate” into the function to find the “y”.Therefore y=-2

\begin{aligned}y=2x^{a}+8x+6\\ y=2\left( -2\right) ^{2}+8\left( -2\right) +6\\ y=2\left( 4\right) -16+6\\ y=8-16+6\\ y=14-16\\ y=-2\end{aligned}

 

Step3: Plug in the value of a, h, k into the vertex form.y=a\left( x-h\right) ^{2}+k.

  • We have a =2, h =-2, k=-2.Therefore plugging these values into the vertex form, we get

\begin{aligned}y=a\left( x-h\right) ^{2}+k\\ y=2\left( x-\left( -2\right) \right) ^{2}+\left( -2\right) \\ y=2\left( x+2\right) ^{2}-2\end{aligned}.

 

ii) Finding X and Y intercepts

  • To find the x-Intercepts,we make y=0 and solve for “x”.

     \begin{aligned}y=2x^{2}+8x+6\\ 0=2\left( x^{2}+4x+3\right) \\ 0=x^{2}+4x+3\\ 0=\left( x+1\right) \left( x+3\right) \\ \Rightarrow x+1=0,x+3=0\\ \Rightarrow x=-1,x=-3\end{aligned}

Therefore the roots of x would be (-1,0) and (-3,0).

  • To find the “y” intercepts we substitute x=0

\begin{aligned}y=2x^{2}+8x+6\\ y=2\left( 0\right) ^{2}+8\left( 0\right) +6\\ y=0+0+6\\ y=6\end{aligned}

Threfore, Y-intercept would be (0,6)

Below is the graph of the y=2x^{2}+8x+6

Graph of 2x^2+8x+6

8) Got a troubling math question?

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No worries…You can post/upload your question in “Free Math Help Forum under a relevant category.

We reply back to your e-Mail with step by step solution.

 

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