# TI-89 zeros, vertex and discriminant?

• Mar 15th 2007, 06:54 PM
power2600
TI-89 zeros, vertex and discriminant?
How do I solve zeros, vertex and discriminant?
I have a TI-89 Titanium.

Y1=(x-4)^2-2
y2=x^2-2x-3
• Mar 24th 2007, 09:30 AM
earboth
Quote:

Originally Posted by power2600
How do I solve zeros, vertex and discriminant?
I have a TI-89 Titanium.

Y1=(x-4)^2-2
y2=x^2-2x-3

Hello,

1. I assume that you have entered the term at y1 and y2 in the (Y=)-editor.

to y1: The equation of this parabola is in vertex form and you can get the coordinates of the vertex: V(4, -2)
To calculate the zeros use the command zeros (F2 + (4)). The correct syntax is shown at the screenshot.
As you can see the discriminant is 2. (The discriminant is the radicand of the root which is added and subtracted from x-value of the vertex. The x-value of the vertex is 4 and to reach the right zero you have to add √(2), to reach the left zero you have to subtract √(2))

to y2: Calculate the zeros. The mean value of the zeros is the x-value of the vertex. So here the vertex is V(1, y2(1)). Plug in the 1 into y2(x) and you'll get -4. Thats the y-value of the vertex: V(1, -4)
The discriminant must be 4 because you have to add √(4) to 1 to reach 3 and you have to subtract √(4) from 1 to reach -1.

If you know calculus you can calculate the vertex by the zeros of the first derivative:
F2+4(F3+1(y2(x),x),x) gives the result {1}

EB