# How to arrange a function quadratics of the graph

At this time I would give a tutorial and the way how to make a graph of graphic function.  This tutorial i ensure will give the best solution for student learning mathematics.

Okey lets we do.

This is the formulas needed to make equation quadratic function.

Case 1

If the graph sketch through the vertex /extreme (p , q)  and any point (x , y). The graph like this

we get the formulas : $\boxed&space;{&space;y-q=a{&space;\left(&space;x-p&space;\right)&space;}^{&space;2&space;}&space;}&space;\\$

Case 2

When the graph of function across X axis we say $\left(&space;{&space;x&space;}_{&space;1&space;},0&space;\right)&space;\quad&space;and\quad&space;\left(&space;{&space;x&space;}_{&space;2&space;},0&space;\right)&space;\\$ , and through any point  $\left(&space;{&space;x&space;},y&space;\right)&space;\\$ then we use this formula :

we get the formula :  $\boxed&space;{&space;y=a\left(&space;x-{&space;x&space;}_{&space;1&space;}&space;\right)&space;\left(&space;x-{&space;x&space;}_{&space;2&space;}&space;\right)&space;}&space;\\$

Lets we see the examples below.

##### Example 1

Identify the graph function of quadratic equation of the picture below.

Alternative Answer

We can see that the graph through the vertex and sure by using the formula case 1

$Vertex:\left(&space;-1,2&space;\right)&space;\quad&space;and\quad&space;any\quad&space;point\quad&space;\left(&space;0,4&space;\right)&space;\\$

$y-q=a{&space;\left(&space;x-p&space;\right)&space;}^{&space;2&space;}\\&space;\\&space;4-2=a{&space;\left(&space;0-\left(&space;-1&space;\right)&space;\right)&space;}^{&space;2&space;}\\&space;\\&space;2=a.1\quad&space;\Longrightarrow&space;\quad&space;a=2$

Substitute (a) to the origin formula then we get

$y-q=a{&space;\left(&space;x-p&space;\right)&space;}^{&space;2&space;}\\&space;\\&space;y-2=2\left(&space;x-\left(&space;-1&space;\right)&space;\right)&space;^{&space;2&space;}\\&space;\\&space;y-2=2{&space;\left(&space;x+1&space;\right)&space;}^{&space;2&space;}\\&space;\\&space;y-2=2\left(&space;x+1&space;\right)&space;\left(&space;x+1&space;\right)&space;\\&space;\\&space;y-2=2\left(&space;{&space;x&space;}^{&space;2&space;}+2x+1&space;\right)&space;\\&space;\\&space;y-2=2x^{&space;2&space;}+4x+2\\&space;\\&space;y=2x^{&space;2&space;}+4x+4$

so the equation of the graph is  $\\&space;y=2x^{&space;2&space;}+4x+4$

##### Example 2

Identify the graph function of quadratic equation of the picture below.

Alternative answer

We can see that the graph across x axis and through  any point and sure by using the formula case 2

$\\&space;intersection\quad&space;of\quad&space;x\quad&space;axis\quad&space;:\quad&space;\left(&space;-4,0&space;\right)&space;\quad&space;and\quad&space;\left(&space;3,0&space;\right)&space;\quad&space;\\&space;any\quad&space;point\quad&space;:\quad&space;\left(&space;0,4&space;\right)$

$\\&space;y=a\left(&space;x-{&space;x&space;}_{&space;1&space;}&space;\right)&space;\left(&space;x-{&space;x&space;}_{&space;2&space;}&space;\right)&space;\\&space;\\&space;4=a\left(&space;0-(-4)&space;\right)&space;\left(&space;0-3&space;\right)&space;\\&space;\\&space;4=a\left(&space;4&space;\right)&space;\left(&space;-3&space;\right)&space;\\&space;4=-12a\quad&space;\Longrightarrow&space;\quad&space;a=-\frac&space;{&space;1&space;}{&space;3&space;}&space;\\$

Substitute (a) to the origin formula then we get

$\\&space;y=a\left(&space;x-{&space;x&space;}_{&space;1&space;}&space;\right)&space;\left(&space;x-{&space;x&space;}_{&space;2&space;}&space;\right)&space;\\&space;\\&space;y=-\frac&space;{&space;1&space;}{&space;3&space;}&space;\left(&space;x-(-4)&space;\right)&space;\left(&space;x-3&space;\right)&space;\\&space;\\&space;y=-\frac&space;{&space;1&space;}{&space;3&space;}&space;\left(&space;x+4&space;\right)&space;\left(&space;x-3&space;\right)&space;\\&space;\\&space;y=-\frac&space;{&space;1&space;}{&space;3&space;}&space;\left(&space;{&space;x&space;}^{&space;2&space;}+x-12&space;\right)&space;\\&space;\\&space;y=-\frac&space;{&space;1&space;}{&space;3&space;}&space;{&space;x&space;}^{&space;2&space;}-\frac&space;{&space;1&space;}{&space;3&space;}&space;x+4\\&space;so\quad&space;the\quad&space;equation\quad&space;of\quad&space;the\quad&space;graph\quad&space;is\quad&space;y=-\frac&space;{&space;1&space;}{&space;3&space;}&space;{&space;x&space;}^{&space;2&space;}-\frac&space;{&space;1&space;}{&space;3&space;}&space;x+4\\$

##### Example 3

Identify the graph function of quadratic equation of the picture below.

Alternative answer

We can see that the graph through the vertex and any point x axis and sure by using the formula case 1

$\\&space;vertex\quad&space;:\quad&space;\left(&space;3,8&space;\right)&space;\quad&space;any\quad&space;point\quad&space;(5,0)\\&space;\\&space;y-q=a{&space;\left(&space;x-p&space;\right)&space;}^{&space;2&space;}\\&space;\\&space;0-8=a\left(&space;5-3&space;\right)&space;^{&space;2&space;}\\&space;\\&space;-8=a\left(&space;2&space;\right)&space;^{&space;2&space;}\\&space;\\&space;-8=4a\quad&space;\Longrightarrow&space;\quad&space;a=-2\\$

Substitute (a) to the origin formula then we get

$\\&space;y-q=a{&space;\left(&space;x-p&space;\right)&space;}^{&space;2&space;}\\&space;\\&space;y-8=-2\left(&space;x-3&space;\right)&space;^{&space;2&space;}\\&space;\\&space;y-8=-2\left(&space;x-3&space;\right)&space;\left(&space;x-3&space;\right)&space;\\&space;\\&space;y-8=-2\left(&space;{&space;x&space;}^{&space;2&space;}-6x+9&space;\right)&space;\\&space;\\&space;y-8=-2{&space;x&space;}^{&space;2&space;}+12x-18\\&space;\\&space;y=-2{&space;x&space;}^{&space;2&space;}+12x-10\\&space;\\&space;so\quad&space;the\quad&space;equation\quad&space;of\quad&space;the\quad&space;graph\quad&space;is\quad&space;y=-2{&space;x&space;}^{&space;2&space;}+12x-10\\$

2 Comments
Follow This :