Am looking for a website which can solve quadratic equations such this
solve for $v$ , $h(v-t) = h(v+t) $ where $h(x) = ax^2 + bx +c$. The value is the vertex of the quadratic function the website wolfarmalpha can solve that when $a,b$ and $c$ are numbers. I want the students to figure the vertex of the quadratic function.
Any ideas? Thanks.
I think Geogebra can handle a particular case for your question on demonstrating the shifting effect.
For the s/w to have something to plot, we give it a sample curve like $h(x) = x^2 - x - 12$.
Next, we ask the s/w to plot y = h(x - 2), where 2 is just particular value for t. The result clearly shows the resultant curve is just a right shift of the original by 2 units.
Plotting y = h(x + 2) is a left shift of the original.
Solving h(x - t) = h (x + t) is to find the point of intersection of the plotted curves.
Since the original quadratic is symmetric (with x = $-b \over {2a}$ being the axis), the two plotted curves are just displaced right and left of the same amount. They will meet at x = $-b \over {2a}$
It is $$h(v-t)=a(v-t)^2+b(v-t)+c$$ $$h(v+t)=a(v+t)^2+b(v+t)+c$$ we get $$-2bt=4avt$$
You have $$a(v-t)^2+b(v-t)+C = a(v+t)^2+b(v+t)+C$$
It simplifies to $$2avt+bt =0$$
Solve for $v$ to get $$v=\frac {-b}{2a}$$
