Functions of the Form <em>y = x<sup>2</sup></em>
Functions of the form \(y={x}^{2}\)
Functions of the general form \(y=a{x}^{2}+q\) are called parabolic functions. In the equation \(y=a{x}^{2}+q\), \(a\) and \(q\) are constants and have different effects on the parabola.
Example
Question
Complete the following table for \(f(x)={x}^{2}\) and plot the points on a system of axes.
|
\(x\) |
\(-\text{3}\) |
\(-\text{2}\) |
\(-\text{1}\) |
\(\text{0}\) |
\(\text{1}\) |
\(\text{2}\) |
\(\text{3}\) |
|
\(f(x)\) |
\(\text{9}\) |
-
Join the points with a smooth curve.
-
The domain of \(f\) is \(x\in \mathbb{R}\). Determine the range.
-
About which line is \(f\) symmetrical?
-
Determine the value of \(x\) for which \(f(x)=6\cfrac{1}{4}\). Confirm your answer graphically.
-
Where does the graph cut the axes?
Substitute values into the equation
\begin{align*} f(x) & = {x}^{2} \\ f(-3) & = {(-3)}^{2} = 9\\ f(-2) & = {(-2)}^{2} = 4\\ f(-1) & = {(-1)}^{2} = 1\\ f(0) & = {(0)}^{2} = 0\\ f(1) & = {(1)}^{2} = 1\\ f(2) & = {(2)}^{2} = 4\\ f(3) & = {(3)}^{2} = 9 \end{align*}
|
\(x\) |
\(-\text{3}\) |
\(-\text{2}\) |
\(-\text{1}\) |
\(\text{0}\) |
\(\text{1}\) |
\(\text{2}\) |
\(\text{3}\) |
|
\(f(x)\) |
\(\text{9}\) |
\(\text{4}\) |
\(\text{1}\) |
\(\text{0}\) |
\(\text{0}\) |
\(\text{4}\) |
\(\text{9}\) |
Plot the points and join with a smooth curve
From the table, we get the following points:
\((-3;9), (-2;4), (-1;1), (0;0), (1;1), (2;4), (3;9)\)Determine the domain and range
Domain: \(x\in \mathbb{R}\)
From the graph we see that for all values of \(x\), \(y \ge 0\).
Range: \(\{y:y\in \mathbb{R}, y\ge 0\}\)
Find the axis of symmetry
\(f\) is symmetrical about the \(y\)-axis. Therefore the axis of symmetry of \(f\) is the line \(x=0\).
Determine the \(x\)-value for which \(f(x)=6\cfrac{1}{4}\)
\begin{align*} f(x)& = \cfrac{25}{4} \\ \therefore \cfrac{25}{4}& = {x}^{2} \\ x& = ±\cfrac{5}{2}\\ & = ±2\cfrac{1}{2} \end{align*}See points \(A\) and \(B\) on the graph.
Determine the intercept
The function \(f\) intercepts the axes at the origin \((0;0)\).
We notice that as the value of \(x\) increases from \(-\infty\) to \(\text{0}\), \(f(x)\) decreases.
At the turning point \((0;0)\), \(f(x)=0\).
As the value of \(x\) increases from \(\text{0}\) to \(∞\), \(f(x)\) increases.
This lesson is part of:
Functions I