SOLUTION: What is the range of y = 1 - x2; Domain = {-1, 0, 1} && y = 4x - x2; Domain = {-1, -2, 0} && y = t2 + t - 2; Domain = {-2, -1, 0, 1} && y = z2 - 3z; Domain = {-1, 0

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: What is the range of y = 1 - x2; Domain = {-1, 0, 1} && y = 4x - x2; Domain = {-1, -2, 0} && y = t2 + t - 2; Domain = {-2, -1, 0, 1} && y = z2 - 3z; Domain = {-1, 0      Log On


   

Question 316831: What is the range of y = 1 - x2; Domain = {-1, 0, 1}
&&
y = 4x - x2; Domain = {-1, -2, 0}
&&
y = t2 + t - 2; Domain = {-2, -1, 0, 1}
&&
y = z2 - 3z; Domain = {-1, 0, 1, 2}
&&& also how exactly do you find the domain like if the problem was y = 2/
x + 3
Thank you!
Katie'
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
+y+=+1+-+x%5E2+
Domain = {-1, 0, 1}
y%28-1%29=1-x%5E2=1-1=0
y%280%29=1-0=1
y%281%29=1-1=0
Range=(0,1)
.
.
.
y+=+4x+-+x%5E2
Domain = {-1, -2, 0}
y%28-1%29+=+4%28-1%29+-+1=-5
y%28-1%29+=+4%28-2%29+-+4=-12
y%280%29+=+4%280%29+-+0=0
Range=(0,-5,-12)
.
.
.
y+=+t%5E2+%2B+t+-+2
Domain = {-2, -1, 0, 1}
y%28-2%29+=+4-2-2=0
y%28-1%29+=+1-1-2=-2
y%280%29+=+0-0-2=-2
Range=(0,-2)
.
.
.
y+=+z%5E2+-+3z
Domain = {-1, 0, 1, 2}
y%28-1%29+=+1+-+3%28-1%29=1%2B3=4
y%280%29+=+0+-+0+=+0
y%281%29+=+1+-+3+=+-2
y%282%29+=+4+-+6=+-2
Range=(4,0,-2)
.
.
.
y+=+2%2F+%28x+%2B+3%29
The domain is all x for which the function makes sense.
For this function, all x are allowed except for the x which makes the denominator equal to zero because division by zero is undefined.
Find the value when the denominator equals zero.
x%2B3=0
x=-3
So then the domain is, in interval notation,
(-infinity,-3)U(-3,infinity)