How do you find the range of f(x) = -x^2 + 3f(x)=x2+3?

3 Answers
Sep 21, 2015

{y|y<=3}{yy3}

Explanation:

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

The number aa 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 yy. 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 kk tells us the minimum/maximum value of yy.

Luckily, the function f(x)=-x^2+3f(x)=x2+3 is already in vertex form. You can look at it this way:
f(x)=-(x-0)^2+3f(x)=(x0)2+3

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

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

The range will then be {y|y<=3}{yy3}. You may also write it in set interval notation as (-oo,3](,3].

Sep 21, 2015

(-oo,3](,3]

Explanation:

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

Sep 21, 2015

I found: -oo<<yy<=33

Explanation:

This function is represented graphically by a downwards parabola; this is because the -11 in front of the x^2x2 term.
To find the range (= possible yy values) we need to find the highest point reached by our parabola, the vertex.
The xx coordinate of the vertex is given as: x_v=-b/(2a)xv=b2a
where the coefficients b and abanda are found writing the function in general form as:
f(x)=ax^2+bx+c=-1x^2+0x+3f(x)=ax2+bx+c=1x2+0x+3
so:
x_v=-(0)/(-1*2)=0xv=012=0
giving for yy (substituting x=0x=0 into your function):
y_v=f(0)=3yv=f(0)=3

So the range (= possible yy values) is:
-oo<<yy<=33

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