SOLUTION: {{{ y= x^2+3 }}} where: -1 ≤ x ≤ 2 Find the minimum, maximum, and range. What I've tried: We already know y = k {{{ f(x)=a(x-h)^2+k }}} so: y = k = 3 N

Algebra ->  Test -> SOLUTION: {{{ y= x^2+3 }}} where: -1 ≤ x ≤ 2 Find the minimum, maximum, and range. What I've tried: We already know y = k {{{ f(x)=a(x-h)^2+k }}} so: y = k = 3 N      Log On


   

Question 303891: +y=+x%5E2%2B3+
where:
-1 ≤ x ≤ 2
Find the minimum, maximum, and range.

What I've tried:
We already know y = k
+f%28x%29=a%28x-h%29%5E2%2Bk+
so:
y = k = 3
Next find the value for x:
+y=ax%5E2%2Bbx%2Bc+
Plug in the variables from the equation:
+y=1x%5E2%2B0x%2B3+
To solve, use:
x= -b/2a
so:
x= -0/2(1)
x=0
Vertex of a graph:
(x,y)
(0,3)
To find the minimum and maximum, take the original equation and calculate using the 3 different variables for x we now have. (-1, 0, and 2; since -1 ≤ x ≤ 2)
+f%28x%29=x%5E2%2B3%0D%0A%7B%7B%7B+f%28-1%29=%28-1%29%5E2%2B3+
= 1 + 3
= 4
+f%280%29=0%5E2%2B3+
= 0 + 3
= 3
+f%282%29=2%5E2%2B3+
= 4 + 3
= 7

Minimum: 3
Maximum: 7
Range: [3,7] = 4

It's what I tried, but apparantly it's wrong. Can anyone help me see where I messed up? Thanks!!!
Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
your answers look okay

the problem states the domain as ___ -1 ≤ x ≤ 2

so maybe the range is ___ 3 ≤ y ≤ 7