Tutoring math, you notice a new technique now and then.  The math tutor shares one he recently heard about.

I first heard of partial factoring over ten years ago.  Two students were discussing it. Since neither asked me about it, I never found out what they meant.  The event recurred every few years.

Partial factoring was never taught to me in school, but I finally got a look at it the other day in someone’s textbook.  It’s a really nice method for finding the vertex of a parabola. Here’s how it goes:

Example:  Find the vertex, axis of symmetry, max or min value, domain, and range of y=-2x² + 6x + 14

Solution: We see that

y=14 whenever -2x^2 + 6x=0

To find the two x-coordinates where y=14, we solve

-2x^2+6x=0

using the common factor technique:

-2x(x-3)=0

which implies that

-2x=0 or x-3=0

and finally

x=0 or x=3

Therefore, y=14 when x=0 or when x=3.

Knowing the shape of a parabola (which y=-2x² + 6x + 14 is), we can draw a rough sketch.

Note the vertex is at the top because the lead coefficient (-2, in this case) is negative. The vertex will be midway between the two x values where y is 14: x=0 and x=3. Getting the average of those two values, we arrive at 1.5, which means that the vertex will be at (1.5, y). To find the y value of the vertex, we put 1.5 for x into the equation:

y=-2(1.5)^2+6(1.5)+14
y=18.5

Therefore, we have (1.5,18.5) as our vertex.

The A/S, which means “axis of symmetry”, is the vertical line x=1.5, as shown in red. The graph’s maximum value is the y coordinate of the vertex: 18.5. The range must therefore be y≤18.5, since the graph goes down from there.  The domain is all real numbers; it always is for quadratic equations, of which y=-2x²+6x+14 is one.

Partial factoring makes finding the vertex of a quadratic function very easy. I’m sure you’ll agree:)

Source: Nelson Foundations of Mathematics 11.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.