How do you find the range of # f(x) = -x^2 + 3#?

3 Answers
Sep 21, 2015

Answer:

#{y|y<=3}#

Explanation:

Since this function is quadratic, its graph is a parabola and it either has a minimum or a maximum value for #y#. To solve for the minimum/maximum value, we convert our equation to the vertex form #y=a(x-h)^2+k# by "completing the square".

The number #a# determines whether the parabola opens upward or downward. This is important because it will tell us whether we're looking for the minimum or maximum value of #y#. If it is positive, then we are looking for the minimum value. If it is negative, we are looking for the maximum value. The number then #k# tells us the minimum/maximum value of #y#.

Luckily, the function #f(x)=-x^2+3# is already in vertex form. You can look at it this way:
#f(x)=-(x-0)^2+3#

First, let's look at #a#. In this equation, #a=-1#. Since it is negative, it means that we are looking for the maximum value of #y#.

Next, we look at #k#. In this equation, #k=3#, meaning 3 is the maximum value for #y#.

The range will then be #{y|y<=3}#. You may also write it in set interval notation as #(-oo,3]#.

Sep 21, 2015

Answer:

# (-oo,3]#

Explanation:

The given function represents a parabola opening downwards with vertex at (0,3). Hence its range would be #(-oo, 3]#

Sep 21, 2015

Answer:

I found: #-oo<##y##<=3#

Explanation:

This function is represented graphically by a downwards parabola; this is because the #-1# in front of the #x^2# term.
To find the range (= possible #y# values) we need to find the highest point reached by our parabola, the vertex.
The #x# coordinate of the vertex is given as: #x_v=-b/(2a)#
where the coefficients #b and a# are found writing the function in general form as:
#f(x)=ax^2+bx+c=-1x^2+0x+3#
so:
#x_v=-(0)/(-1*2)=0#
giving for #y# (substituting #x=0# into your function):
#y_v=f(0)=3#

So the range (= possible #y# values) is:
#-oo<##y##<=3#

Graphically:
graph{-x^2+3 [-10, 10, -5, 5]}