Math Revision 15.2 - Quadratic Graphs Part Two

In these revision notes for Quadratic Graphs Part Two, we cover the following key points:

• How do we use simple factorization to plot a quadratic graph?
• How do we identify the relevant points needed to plot a quadratic graph?
• How do we plot a quadratic graph by identifying the numbers using the "Completing-the-Square" method?
• How do we make the distinction between various cases in quadratic graphs?
• How do we express new coefficients in the "Completing-the-Square" method in terms of the original ones?
• How do we plot a quadratic graph using the "Completing-the-Square" method?

Quadratic Graphs Part Two Revision Notes

The quadratic equations with one variable

can be solved by factorising the left part to show it as a product of two expressions in the brackets. This can be done only in two cases out of three possible.

Case 1: When the quadratic equation has two distinct roots (i.e. when the discriminant Δ is positive). In this case, the above equation is written in the form

where

and

When the task is to plot the graph of a quadratic equation with two variables

we ignore the existence of variable y first and try to factorise the part containing variable x according to the above rules. The factorisation allows us to find the x-intercepts, as each solution corresponds to an x-intercept. We, therefore, write the factorised form of a quadratic equation with two variables as

and when solving it, we take y = 0 and write

where x₁ and x² are the solutions (roots) of the quadratic equation with one variable, which correspond to the x-intercepts of the same quadratic equation but with two variables when shown in a graph.

The rest of the points are identified in the same way as when using the quadratic formula to find the roots.

Case 2: When the quadratic equation has a single root (when the discriminant Δ is zero). In that case, the part containing the independent variable can be written as a binomial expression, so the equation is written in the form

where m is a coefficient and n is a constant.

The corresponding quadratic equations with one variable of the type

have a single solution (root) of the form

Since the original form of the quadratic equation with two variables is

and given that we can express the new factorised form as

then we have

In this way, we can find the coefficient m and the constant n if we are convinced that the original quadratic equation has a single root. These values help finding the only root of the equation, which also gives the vertex V of the parabola. The rest of the solution is the same as in the previous methods.

Case 3: When the discriminant is negative, there are no x-intercepts at all, so, it is impossible to factorise the expression containing the independent variable. Therefore, we use the same procedure as above to find the necessary points to sketch the graph. Thus, first, we find the vertex V, then the point C that corresponds to the y-intercept, then its symmetrical point D, and at the end, two equidistant points from the vertex: one on its left and the other on its right side. Then, we plot the graph based on these five points.

Some quadratic equations with one variable can be solved by completing the square according to the formula

where a = 1, p = b/2 and q is the remaining part of the original constant c.

What if the coefficient a is different from 1? Well, in this case, we must try to write the simplified version of the quadratic equation (without the variable y)

in the form

where

In this way, we obtain

After expressing the original quadratic equation with two variables in the form

to complete the square, we find the roots (if any) that indicate the x-intercepts and then, we find the rest of the points as in the other cases discussed so far.