Vertex

boxedexe

I have a problem with something. I have points, (20,9.1), (40,7.8), (60,7.1), (80,7.1), (100,7.8), and (120,9.1). These are the only information given to me. How would I be abel to figure the vertex of this quadratic line with teh tools of calculus?

Masoud Zargar

**Posted:** Wed Oct 12, 2005 10:54 pm

jmerry

It doesn't take any calculus at all; just look at the symmetry and see that it must be at $x=70$ . The $y$ -coordinate is a little trickier, as we're interpolating with inexact values. Still, it's pretty easy to see that that value is about 7.0.

**Posted:** Wed Oct 12, 2005 11:27 pm

boxedexe

Well, it is obvious that $x=70$ , however, how would I be able to figure out the y-coordinate without graphing?

Masoud Zargar

**Posted:** Wed Oct 12, 2005 11:30 pm

jmerry

$(7.8-y)=a\cdot30^2$
$(7.1-y)=a\cdot10^2$
$\frac{7.8-y}{7.1-y}=9$
$.7=8\cdot(7.1-y)$
$y=7.0$

All of this is of course to the number of significant digits we have, which is small.

**Posted:** Wed Oct 12, 2005 11:38 pm