Range and Domain of Quadratic Functions

I know two ways to solve range and domain of quadratic functions. First is the graphical method and the second one is the algebraic method. This topic will just focus on quadratic functions, which means that the opening is either up or down.

Consider the figure:

drquadGraphically, the range of quadratic function is from the y-coordinate to positive infinity (if the graph opens upward) or negative infinity (if the graph opens downward). The domain is the set of all real numbers always since it extends from left to right.

The second way is to solve it algebraically. Given a quadratic function,


We solve for y in terms of x

Rearrange the term and equate all terms to 0.


Treat y as constant. By quadratic formula we can solve for x in terms of y now.

x=\displaystyle\frac{-b\pm\sqrt{b^2-4ac}}{2a}    where, a=a, b=b, c=c-y

By direct substitution,

x=\displaystyle\frac{-b\pm\sqrt{b^2-4a(c-y )}}{2a}

To solve for range,

b^2-4a(c-y)\geq for it to become real.

Solving for y,


What a surprise! Isn’t it the same formula to get the y-coordinate of the vertex of quadratic equation?


Worked Example 1: 2007 MMC divison finals

Find the range of y=9-x^2



Algebraic Method:

Solve for x in terms of y.

x^2=9-y x=\sqrt{9-y} 9-y\geq0

y\leq 9 or in symbols (-∞,9]


Graphical Method:

Again get the y-coordinate of the vertex using the formula

y=\displaystyle\frac{4ac-b^2}{4a} but b=0.

The formula can be simplified to

y=c , the next step is look at the sign of a if the graph is opening downward or upward. Since a here is negative the graph opens downward. Therefore the range is y\leq9 or (-∞,9]


Sample Problem 2:

Find the range of the function f(x)=3x^2-2x+4



y=\displaystyle\frac{4ac-b^2}{4a} y=\displaystyle\frac{4(3)(4)-(-2)^2}{4(3)}

y=\displaystyle\frac{11}{3} and the graph is opening upward.

The range is y\geq\displaystyle\frac{11}{3} or [\displaystyle\frac{11}{3},+\infty)


If you have questions or you want to add something feel free to leave a comment below.


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