<p>Hello :)</p>
<p>I was doing Barrons exercise from parametric functions.</p>
<p>when i graphed this parametric function in ti 89 in the parametric mode,it showed me answer x> 0
not x> 1/4 which is actually answer.</p>
<p>equation:</p>
<p>x=t^2+t
y=t^2-t</p>
<p>sorry,the right answer was:</p>
<p>x> -1/4</p>
<p>So it seems that what you’re being asked to do, is minimize the function x=t^2+t for any t.</p>
<p>There are two approaches to this:</p>
<p>If you know calculus already, just take the derivative and set it equal to zero.
2t+1=0
t=-1/2
which means min x = -1/4 [plug in -1/2 for t]</p>
<p>If you don’t know calculus, use the formula for the vertex of a parabola, which also happens to be the minimum of upwards-opening parabolas like this one.</p>
<p>t coordinate is -b/2a = -1/(2*1) = -1/2
so min x = -1/4</p>
<p>Does that help?</p>
<p>For parametric functions i thought i would use my graphic calculator.
is not it possible to do it just by graphing it and observing?</p>
<p>also in the book they graphed this function but it was different from what i graphed.
In the function they graphed it was very easy to see that x>-1/4</p>
<p>Thank you</p>
<p>Certainly you can graph it, though unless you zoom in with a really well chosen viewing window it might be hard to see where the minimal x value is, which is why I prefer the computational way (which is really quick in any case).</p>
<p>If you zoom in around the point (x,y) = (-1/4, 3/4) do you see how x reaches a minimum there?</p>
<p>SarumanTheWise is a wise guy:)</p>
<p>Thanks a lot!</p>
<p>your suggestion motivated me to search in Google about parametric functions solving with TI 89 .
Maybe other students with the same probleml benefit from this thread!</p>
<p>The only thing you should do is Reset the Window values to
tmin = -3 xmin = -3 ymin = -3
tmax = 3 xmax = 3 ymax = 3
tstep = 0.1 xscl = 1 yscl = 1</p>
<p>It will be nice if it does not change the numbers in windows and save like that:)</p>