## Learning objectives:

This lesson focuses primarily on graphing quadratic functions, and how can vertex of the parabola help in easy parabola graphing.Other supporting topics include quadratic parent function, leading coefficient “a”, and Transformations.

The following is the brief of key concepts and  of this article:

• Graphing quadratic functions using the conventional approach.
• The best strategy for easy graphing of Quadratic function.
• Summary of steps for graphing quadratic functions using the vertex.
• Worked out example.
• Free Math Help forum

The graph of a quadratic function is called as a parabola.

A quadratic function is a non-linear function with the highest degree of the variable as 2.For graphing a linear function, all we need is 2 points.

But for graphing a non-linear function we need more than two points. In fact, more the points, better would be the graph when it comes to graphing of a non-linear function.

ii)Table of ordered pairs

The standard approach of graphing a parabola would be by creating a table of ordered pairs.

we can select few arbitrary values for the”x” -coordinates of our choice and can find the corresponding”y” coordinates.

Once we find the  “y” coordinate, corresponding to the chosen “x” -coordinate, we get ordered pair (x,y) for plotting on the graph.

After having a good number of ordered pairs found, we can plot the points on the graph and can join the points together to give a smooth parabolic graph.

Often, we are inclined towards picking random values of “x” of our choice to find the coordinate points for graphing quadratic functions. This strategy might not be of excellent help all of the time.

Sometimes, we land up in getting sprinkled coordinate points which would be a challenging task to complete the quadratic function graphing.

There is every chance that the vertex of the parabola is not captured properly in this approach.

As shown below is the table of ordered pairs for the graph of y =x^2 with few values of “x” taken at random.

Prime Student 6-month Trial This leads to sprinkled points all over the graph. When we join the points smoothly, we get an incorrect impression about the parabolic graph, especially the vertex of y =x^2.

## 2. What’s the best approach for graphing a quadratic function?

Find the vertex of the parabola:

The perfect method to start graphing quadratic functions is by discovering the vertex of the quadratic.

As discussed in the prior course, the vertex is the turning point of the parabola and it is the point where the axis of symmetry intersects the parabola. It is the minimum or a maximum of a parabola.

Consider the quadratic function . The x-coordinate of the vertex “h” is given by the formula h = – b/2a.

We can strategically use the vertex point, for the collection of the pair of  “x”-coordinates which are required for the smooth graphing of the quadratic function.

All we need is to pick up the “x” coordinates which are equidistant from the vertex.

Choosing the equidistant  (or) symmetrical points about the  vertex:

For, instance, Let us consider the quadratic y=x^2 whose vertex (h,k) is located at (0,0). So we can say that h=0 and k=0.

It is evident that x= -1 and x=1 are equidistant from the vertex h=0.When x=1 and -1 corresponding y coordinates for  x =1 and -1 would be 1.

Therefore the corresponding points on the graph  (-1,1) and (1,1) will be symmetrical about the axis of symmetry of y= x^2 which in turn passes through the vertex (0,0) of the quadratic.

Therefore, we can choose this pair of points to fill out the table of coordinate points.

Likewise, we can pick the other pair of”x”-Coordinates which are at equidistance from h =0 like x =2 & – 2, x =3 & -3 to fill the  table of coordinate points.

As a result, we will be able to select the pair of “x” coordinates, by avoiding scattered points, to complete the Table of ordered pairs.

The key advantage of this approach is that finding the “y” coordinates to one-half of the symmetric points is adequate.

If we find the “y” coordinates of the function y=x^2  corresponding to say x =1,2,3(which are 1,4,9),

the “y” coordinates corresponding to x= -1,-2,-3 would also be  y=1,4,9 as shown in the picture below.

Therefore the coordinate pairs can be  (0,0) { (1,1),(-1,1)} ; { (2,4),(-2,4) } ;{3,9),(-3,9)} which are symmetrical about the “y” axis. Here is the summary of the steps for graphing the quadratic functions using the vertex.

## 3) Strategy Step By Step for graphing quadratic functions using vertex.

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

Step2: Find the vertex (h,k)of the Quadratic using the formula  h=-b/(2a) using “h” find the value of “k” thereby.

Step3: Choose the pairs of “x” coordinates  which are equidistant from “h”(say {x1, x1′} ,{ x2,x2′},{x3,x3′}…)

Step4: Find the corresponding “y” coordinates (y1 ,y2,y3..) for “x” coordinates of each pair(x1,x2,x3…) giving coordinates (x1,y1) ,(x2,y2) , (x3,y3) …

Step5: Just repeat the “y” coordinates  already found (y1,y2,y3..), to the corresponding left out  “x” coordinate (x1′ ,x2′,x3’…)giving raise to coordinate points (x1′,y1),(x2′,y2),(x3′,x3)..

Step6: Plot these points on the graph and join them with a smooth parabolic shaped curve.

Let’s look at an example problem to better understand the steps.

## 4. Worked out examples

i) Graph the quadratic function.y= x^2 – 4x

#### Strategy Step By Step for graphing quadratic functions

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

Given x^2 – 4x.Therefore we can say that coefficient a =1, b = -4  and c=0.

Step2: Find the vertex (h,k)of the Quadratic using the formula  h=-b/(2a) using “h” find the value of “k”. thereby.

We can observe that the value of .

The corresponding value of “k” would be Step3: Choose the pair of “x” coordinates equidistant from the “h” (vertex’s X-coordinate).

Let us pick the pair of numbers equidistant from “h” =2 by 1 unit. The corresponding pair of numbers would be (2-1)=1,(2+1)=3 Therefore we can pick this pair of “x” coordinates{1,3} for graphing.

Similarly, let us choose a pair of x-coordinates which are equidistant from the vertex by  2  units Therefore we can have x -coordinates as (2+2) =4, (2-2) =0, and the new pair would be {0,4}.

Likewise, the pair of x-coordinates equidistant from h =2 by 3 units would be (3+2)=5 and (2-3)=-1 the pair as

So we now  have  3  pairs of x -coordinates equidistant from h=2 which  are {1,3};{0,4};{-1,5}

Step4: Choose 3 numbers, one from each pair, from the three available pairs of x- coordinates. Find the corresponding “y” coordinates for it.

Let us choose the x=-1, 0,1  respectively from each pair since they are of small magnitude and it is evaluating the function.

The table below shows the “y” values corresponding to the “x” values – 1,0,1.

We can observe that the parabola graphing is half completed with these 3 points and vertex. Step5: Make the ordered pairs by Just repeat the “y” coordinates already found, the corresponding left out x -coordinates numbers in each pair.

We found that for “x” values { -1,0,1} corresponding “y” values are 5,0,-3 respectively  .

The left out numbers in each pair of x -coordinates would be  3,4,5 and we need not calculate for them.

Therefore the left out x values and corresponding y coordinates make 3 more points which would be (3,-3),(4,0),(5,5).

Step6: Plot these points on the graph and join them by a smooth curve to get a decent parabola curve.

Since  we have vertex as (2,-4) and symmetrical points of the quadratic as  (1,-3)(3,-3) ;(0,0)(4,0);(-1,5)(5,5) we can plot these points and draw a smooth curve joining these points making a parabola. Therefore for complete graphing of a quadratic, vertex and one-half of the symmetric points are sufficient.

## 5.FREE Math Help forum

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#### 1 Comment

1. Eulalia

It works quite well for me