Question 706046
find the x and y intercepts and the range of the following: 
------------------------------
f(x)=x^2-6x+4
----
y-int:
Let x = 0, then y = 4
------
x-int:
Let y = 0,
x = [6 +- sqrt(36-4*1*4)]/2
x = [6 +- sqrt(20)]/2
x = [6 +- 2sqrt(5)]/2
x = [3+-sqrt(5)]
--------------------------
To find the range, find the vertex:
vertex occurs at x = -b/(2a) = 6/2 = 3
f(3) = 3^2-6*3+4 = 9-18+4 = -5
-----
So, you have a parabola opening upward from the
point (3,-5)
Range = All Real Numbers >= -5
====================================
Cheers,
Stan H.
===================