SOLUTION: how can you find the domain and range of y=-3/4(x-1/2)^2+9 ?

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Question 856008: how can you find the domain and range of y=-3/4(x-1/2)^2+9 ?
Found 2 solutions by KMST, josgarithmetic:
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
It is a quadratic function.
As for all polynomial functions,
its domain is all the real numbers.
Because it is a quadratic function
(one of those that graph as parabolas),
its range is more limited.
Quadratic functions have a minimum (if the coefficient of the term with x%5E2 is positive),
or a maximum (when that coefficient is negative).
In this case the coefficient, -3%2F4 , is negative, and there is a maximum.
You do not need to know anything about quadratic functions to know that the square of a number is at least zero (if that number is zero), and often positive, but never negative.
%28x-1%2F2%29%5E2 is positive or at least zero: %28x-1%2F2%29%5E2%3E=0 .
So, -%283%2F4%29%28x-1%2F2%29%5E2 is negative or at most zero: -%283%2F4%29%28x-1%2F2%29%5E2%3C=0 .
As a consequence, y=-%283%2F4%29%28x-1%2F2%29%5E2red%28%22%2B+9%22%29 is at most 9 :
y=-%283%2F4%29%28x-1%2F2%29%5E2%2B+9%3C=9
The range includes only 9 and all real numbers that are less than 9.
Some would write is as (-infinity,9].
Others may prefer a different notation, such as {x | x%3C=9}

Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
Parabola in standard form. y as a function of x%5E2, so axis of symmetry is parallel to the y axis.
DOMAIN: all real numbers.
RANGE: Coefficient is negative on the x%5E2 term and the vertex is therefore a maximum, at (1/2, 9). Now, range?