Geogebra is really useful in investigating the shape of a quadratic. If we start with y=ax² + bx + c, and let geogebra create the slider (which default to 1) we get the following image:
Students from Year 11/10 might know how varying a and c will affect the parabola. But what about b? Many maths teachers I’ve spoken to find this difficult to answer and the result of varying b is perhaps surprising. Dynamic geometry software enables us to see the effect of this in a quick, precise and powerful manner. It is possible to trace the curve of y=ax² + bx + c as b varies, and this results in a rather ominous black batman shape, so I suggest the following activity which brings in a nice bit of differentiation concerning stationary points. (Although asking why does the curve always go through (0,1) is worthwhile)
Let’s begin with y=x²+bx. By setting a=1 and c=0 we can hone in on the effect of varying b. We can identify the stationary point of the curve by using the extremum/turning point command on Geogebra.
The image below shows the trace of the stationary point whilst varying b. This can (and should!) be proved algebraically and is a nice activity for Year 12.
y = x²+bx
dy/dx = 2x + b = 0
b = -2x
Now sub this into the original to get;
y = x² + (-2x)x
y = -x²
You can repeat the above but vary a instead. I recommend starting with y = ax^2+x and tracing the stationary point as before. A similar algebraic approach will yield the required equation of the path of the trace of the stationary point (y = 0.5x)
I’ll leave you to do the same whilst varying c. It may not be a surprise that you’ll find the equation of the path of the trace of the stationary point is x = 0 if you start with y = x^2+c
Update 12th June:
Thanks to Tom Button @mathstechnology for linking the above with completing the square. Changing the language of stationary point to vertex might be enough for students to utilise completing the square as a method to find this point. So, if we complete the square;
y=x²+bx+c
y=(x+b/2)²+c-(b/2)²
the vertices of these are at the points (-b/2, c-(-b/2)²), i.e. as you vary b this traces the points where y=c-x²
This is in agreement with the result proved earlier that the trace of the stationary point is y = -x² as we began with y = x² + bx and so c = 0.
