SOLUTION: Given the function y = f(x) = -x2 + 8x – 12. Suppose the domain of the function is restricted to the semi-open interval (3, 7]. What is the corresponding range of the function in

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Question 1054882: Given the function y = f(x) = -x2 + 8x – 12. Suppose the domain of the function is restricted to the semi-open interval (3, 7]. What is the corresponding range of the function in interval notation?
a. (-5, 4)
b. [-5, 4]
c. [-5, 3]
d. [-5, 3)
e. [-4, 5]

Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
the equation is y = -x^2 + 8x - 12.

the domain is (3,7]

this means that x is > 3 and x <= 7

this can be written as 3 < x <= 7

in interval notation, that looks like (3,7].

the value of y is defined when x > 3 and when x <= 7.

the function is y = -x^2 + 8x - 12.

this is a quadratic function and has a maximum or minimum value at x = -b/2a.

since a = -1 and b = 8 and c = -12, then:

x = -b/2a = -8/-2 = 4.

the function has a maximum value at x = 4.

when x = 7, the value of the function is y = -5.
when x = 4, the value of the function is y = 4

the range of y goes from the max of y to the min of y, within the confines of the domain.

this means that y goes from 4 to -5.
put the minimum value ahead of the maximum value and this means that y goesx from -5 to 4.

x = 4 and x = 7 are both valid values within the domain.
therefore y = 4 and y = -5 are both valid values within the range.

this means that y is >= -5 and y is <= 4.
this can be written as -5 <= x <= 4.

in interval notation, this can be written as [-5,4]

that would be selection b.

if you graph the equation, you will see that this makes sense.

here's the graph.

$$$

even though it's shown on the graph, the coordinate point of (3,3) is not a valid point of the equation because x = 3 is not part of the domain.










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