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Graphing Quadratic Equations Using Factoring

A quadratic equation is a polynomial equation of degree 2 .  The standard form of a quadratic equation is

0 = a x 2 + b x + c

where   a , b and c are all real numbers and a 0 .

If we replace 0 with y , then we get a quadratic function

      y = a x 2 + b x + c

whose graph will be a parabola .

The points where the graph intersects the x -axis will be the solutions to the equation, a x 2 + b x + c = 0 . That is, if the polynomial a x 2 + b x + c can be factored to ( x p ) ( x q ) , we know by the zero product property that if ( x p ) ( x q ) = 0 , either ( x p ) = 0 or ( x q ) = 0 . Then p and q are the solutions to the equation a x 2 + b x + c = 0 and therefore the x -intercepts of the quadratic equation.

Since the x -coordinate of the vertex of a parabola is exactly the midpoint of the x -intercepts , the x -coordinate of the vertex will be p + q 2 .

You can use the x -coordinate of the vertex to find the y -coordinate.

Now you have the vertex and 2 other points on the parabola (namely, the x -intercepts). You can use these three points to sketch the graph.

Example 1:

Graph the function y = x 2 8 x + 12 using factoring.

Compare the equation with the standard form, y = a x 2 + b x + c . Since the value of a is positive, the parabola opens up.

Factor the trinomial, x 2 8 x + 12 . Identify 2 numbers whose sum is 8 and the product is 12 . The numbers are 2 and 6 . That is, x 2 8 x + 12 = ( x 2 ) ( x 6 ) .

x 2 8 x + 12 = 0 ( x 2 ) ( x 6 ) = 0

So, by the zero product property, either ( x 2 ) = 0 or ( x 6 ) = 0 . Then the roots of the equation are 2 and 6 .

Therefore, the x -intercepts of the function are 6 and 2 .

The x -coordinate of the vertex is the midpoint of the x -intercepts. So, here the x -coordinate of the vertex will be 2 + 6 2 = 4 .

Substitute x = 4 in the equation y = x 2 8 x + 12 to find the y -coordinate of the vertex.

y = ( 4 ) 2 8 ( 4 ) + 12 = 16 32 + 12 = 4

That is, the coordinates of the vertex are ( 4 , 4 ) .

Now we have 3 points ( 4 , 4 ) , ( 2 , 0 ) and ( 6 , 0 ) which are on the parabola. Plot the points. Join them by a smooth curve and extend the parabola.

Example 2:

Graph the function y = x 2 2 x + 8 using factoring.

Compare the equation with the standard form, y = a x 2 + b x + c . Since the value of a is positive, the parabola opens up.

Factor the trinomial, x 2 2 x + 8 .

First, factor out 1 .

x 2 2 x + 8 = 1 ( x 2 + 2 x 8 )

Factor the expression in the parenthesis. Identify 2 numbers whose sum is 2 and the product is 8 . The numbers are 4 and 2 . That is, x 2 + 2 x 8 = ( x + 4 ) ( x 2 ) .

Then, the given function becomes y = ( x + 4 ) ( x 2 ) .

So, y = 0 implies, by the zero product property, x + 4 = 0 or x 2 = 0 .

Therefore, the x -intercepts of the graph are 4 and 2 .

The x -coordinate of the vertex of a parabola is the midpoint of the x -intercepts. So, here the x -coordinate of the vertex will be 4 + 2 2 = 1 .

Substitute x = 1 in the equation y = x 2 2 x + 8 to find the y -coordinate of the vertex.

y = ( 1 ) 2 2 ( 1 ) + 8 = 1 + 2 + 8 = 9

So, the coordinates of the vertex are ( 1 , 9 ) .

Now we have 3 points ( 1 , 9 ) , ( 4 , 0 ) and ( 2 , 0 ) which are on the parabola. Plot the points. Join them by a smooth curve and extend the parabola.