As we have seen, a quadratic function is of the form $f(x)=ax^2+bx+c$f(x)=ax2+bx+c where $a$a, $b$b and $c$c are constants, called coefficients.
A quadratic function can be defined over the set of real numbers or over some subset of them, we call this the domain.
The range of a quadratic function depends on the particular coefficients of the function and on the domain. If the function has been defined over its natural domain, there is always either a maximum or a minimum function value. At the opposite extreme, the range extends to either $-\infty$−∞ or $\infty$∞, meaning there is no bounding value apart from the maximum or minimum.
When looking at parabolas, we will most often find that the domain is all real $x$x, also written as the set $\left(-\infty,\infty\right)$(−∞,∞), and that the range is either $\left[\text{min value},\infty\right)$[min value,∞) for concave up parabolas, or $\left(-\infty,\text{max value}\right]$(−∞,max value] for concave down parabolas. This is illustrated in the following image.
However, if the quadratic function is defined over a restricted domain such as an interval, the range also will be restricted. The set of function values will be bounded above and below in this case and some work will be required to determine exactly what the range is.
The function $f$f maps the interval $\left[-4,4\right]$[−4,4] to the real numbers according to the rule $f(x)=x^2-3x+2$f(x)=x2−3x+2. This means that the domain of the function is all the real numbers between (and including) $-4$−4 and $4$4. What is the range of $f$f?
Firstly we check the end points of the interval. We are looking for either a maximum or minimum value. So, we check
$f(-4)$f(−4) | $=$= | $(-4)^2-3\times(-4)+2$(−4)2−3×(−4)+2 |
$=$= | $16+12+2$16+12+2 | |
$=$= | $30$30 |
and
$f(4)$f(4) | $=$= | $4^2-3\times4+2$42−3×4+2 |
$=$= | $16-12+2$16−12+2 | |
$=$= | $6$6 |
Inspection of the graph of the function shows that it has a minimum value at $x=\frac{3}{2}$x=32. At this point $f\left(\frac{3}{2}\right)=\left(\frac{3}{2}\right)^2-3\times\left(\frac{3}{2}\right)+2=-\frac{1}{4}$f(32)=(32)2−3×(32)+2=−14.
We can confirm this algebraically by writing
$f(x)$f(x) | $=$= | $x^2-3x+2$x2−3x+2 |
$=$= | $x^2-3x+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2+2$x2−3x+(32)2−(32)2+2 | |
$=$= | $\left(x-\frac{3}{2}\right)^2-\frac{1}{4}$(x−32)2−14 |
Since the squared term can never be less than $0$0, we see that the minimum occurs at $x=\frac{3}{2}$x=32 and at this point the function value is $-\frac{1}{4}$−14.
So, the range of the function given by $x^2-3x+2$x2−3x+2 over the interval $[-4,4]$[−4,4] is $\left[\frac{-1}{4},30\right]$[−14,30]. The graph is displayed below.
Recall that a function is a mapping from one set to another. We are often concerned with sets of numbers. A function rule tells us what number in the second set is associated with each number in the first set. The first set, the set of numbers that the function rule is applied to, is called the domain of the function.
A function rule is often given in the form of an equation. For example, $y=5x-4$y=5x−4 or $y=x^2-3x+2$y=x2−3x+2. In all similar cases, the function rule is the expression found on the right-hand side of the equation. It specifies what is to be done to the variable $x$x. So, the domain of the function is the set of all possible values of $x$x to which the function rule can be applied.
In the two examples given, the rules make sense for every real number and we say the natural domain of each of these functions is the set of real numbers. However, we can, if we wish, restrict the domain to some smaller set of numbers. For example, we may only be interested in what happens to the numbers in the interval $[-4,4]$[−4,4] and so, we might specify this interval as the domain of the function.
If we apply the function rule to all the numbers in the domain, the result is another set of numbers that we call the range of the function. In the case of the function defined by the equation $y=x^2-3x+2$y=x2−3x+2, the function values $y$y belong to the set of real numbers but, more specifically, they form the interval $[-\frac{1}{4},\infty)$[−14,∞). So, this interval is the range of the function (we explain in the example below how this range interval is found).
The function rule is often described separately from an equation involving the range variable $y$y. We may speak of a function $f$f whose domain is the interval $[-4,4]$[−4,4] given by the rule $f(x)=x^2-3x+2$f(x)=x2−3x+2. The symbol $f(x)$f(x) has the meaning, "this rule gives the value of the function at the domain value $x$x".
Consider the graph of the function $y=f\left(x\right)$y=f(x) and answer the following questions.
What is the $y$y-value of the absolute minimum of the graph?
Hence determine the range of the function.
$y\ge\editable{}$y≥
What is the domain of this function?
All real $x$x
$x\ge-4$x≥−4
$x\le-2$x≤−2
$x>-2$x>−2
All real $x$x
$x\ge-4$x≥−4
$x\le-2$x≤−2
$x>-2$x>−2
Consider a quadratic function that passes through the points shown in the following table.
$x$x | $y$y |
---|---|
$1$1 | $17$17 |
$2$2 | $7$7 |
$3$3 | $1$1 |
$4$4 | $-1$−1 |
$5$5 | $1$1 |
$6$6 | $7$7 |
$7$7 | $17$17 |
Does this quadratic function have a maximum or minimum point?
Maximum
Minimum
Maximum
Minimum
What is the minimum value of the function?
What is the range of the quadratic function? Express the range as an inequality in terms of $y$y.
Consider the parabola defined by the equation $y=x^2+5$y=x2+5.
Is the parabola concave up or concave down?
Concave up
Concave down
Concave up
Concave down
What is the $y$y-intercept of the parabola?
What is the minimum $y$y value of the parabola?
Hence determine the range of the parabola.
$y\ge\editable{}$y≥
Display the graphs of linear and non-linear functions and connect the structure of the functions with their graphs
Apply graphical methods in solving problems