\chapter{Polynomial and rational functions}
\section{Quadratic functions}
A \textit{quadratic function} is a function obtained by transforming the basic parent function
\(y=x^2\) using shifts, reflections and stretching/compressing. So for example the
functions \(y=2(x-1)^2-3\) and \(y=-2x^2+5\) are quadratic functions.
\subsection{Standard form}
In general, if
we stretch or compress \(y=x^2\)
vertically by \(a\),
and then we shift horizontally by \(h\) and vertically by \(k\), we obtain the
\textit{standard form} of a quadratic function:
| Standard form of a quadratic function |
|---|
| \(y=a(x-h)^2+k\) |
Since we shifted the vertex that was originally at \((0,0)\) horizontally by \(h\)
and vertically by \(k\), the vertex of the quadratic function \(y=a(x-h)^2+k\)
is \(V=(h,k)\).
| Vertex of a quadratic function \(y=a(x-h)^2+k\) |
|---|
| \(V=(h,k)\) |
The graph of a quadratic function is called a \textit{parabola}. If \(a\) is a positive number,
then there is no reflection, and we say that the parabola \textit{opens up}
(just like in the original graph of \(y=x^2\) ). But if \(a\) is negative, then the graph is
reflected across the \(x\)-axis, and the parabola
\textit{opens down}.
The figure below shows the graphs of three quadratic functions with \(a > 0\):
\[
\begin{array}{ccc}
\img{U2_1F1.jpg}{}{10em}{} & \img{U2_1F2.jpg}{}{10em}{}
& \img{U2_1F3.jpg}{}{10em}{}\\
y=(x+1)^2+1 & y=2x^2-2 & \displaystyle{y=\frac{1}{2}(x-1)^2}
\end{array}
\]
and the next figure shows the graphs of three quadratic functions with \(a< 0\):
\[
\begin{array}{ccc}
\img{U2_1F4.jpg}{}{10em}{} & \img{U2_1F5.jpg}{}{10em}{}
& \img{U2_1F6.jpg}{}{10em}{}\\
y=-(x+1)^2+2 & y=-2x^2+3 & \displaystyle{y=-\frac{1}{2}(x-1)^2}
\end{array}
\]
Note that a quadratic function always has a \(y\)-intercept, but it may have one, two,
or no \(x\)-intercepts.
\subsection{Finding a parabola from a point and the vertex}
If we are given two points, we can find the straight line that goes through them.
This is no longer true for parabolas: given two points, in general there will be infinitely many
parabolas going through them. But if one of the two points is the vertex, then there
will be only one parabola going through the two points, as explained in the following example.
\begin{example}
Suppose we want a parabola to have the vertex at the point \((2,3)\). Then we already know that the
coordinates \((h,k)\) of the vertex are \(h=2\) and \(k=3\), and so the standard form will be
\(f(x)=a(x-2)^2 + 3\). We still do not know \(a\). But if we also know that the parabola must
go through a given point, for example, the point \((4,5)\), then we can substitute \(x\) with
4 and \(f(x)\) with 5 and solve for \(a\):
\[
\begin{array}{rcll}
5 & = & a(4-2)^2 +3 & \mbox{put \(x=4\), \(f(x)=5\) in the equation}\\[1ex]
5 & = & a(2)^2 +3& \mbox{simplify}\\[1ex]
5 & = & 4a+3 & \\[1ex]
5 \color{red}{-3} & = & 4a + 3 \color{red}{-3} & \mbox{solve for \(a\)}\\[1ex]
2 & = & 4a & \\[1ex]
a & = & \displaystyle \frac{1}{2}
\end{array}
\]
So we can now write the equation of the parabola:
\[f(x)=\frac{1}{2}(x-2)^2 +3.\]
\end{example}
\subsection{Drawing a quadratic function}
We now discuss how to draw an accurate graph of a quadratic function.
The vertex, \(y\)-intercept and any \(x\)-intercepts should always be shown on the graph.
It is good practice (especially if there are no \(x\)-intercepts)
to choose a few extra points to be plotted on both side of the vertex.
As for any function, finding the \(y\)-intercept is easy: we just need to set \(x=0\) in the equation.
To find the \(x\)-intercepts (if any), we need to solve the equation \(f(x)=0\) for \(x\). This may not be easy at all
for some functions. But it is easy for quadratic equations in standard form, as the next example shows.
\begin{example}
Consider the function \(f(x)=2(x+1)^2-8\). We recognize this to be a quadratic function
in standard form, with \(a=2\) (so the parabola opens up), \(h=-1\), \(k=-8\). So the
vertex is \(V=(-1,-8)\). The \(y\)-intercept is found by computing \(f(0)=2(0+1)^2-8=2(1)-8=-6\).
So the \(y\)-intercept is the point \((0,-6)\).
The \(x\)-intercepts are found by solving \(f(x)=0\):
\[
\begin{array}{rcll}
2(x+1)^2-8 & = & 0 \\[1ex]
2(x+1)^2 & = & 8\\[1ex]
(x+1)^2 & = & 4\\[1ex]
x+1 & = & \pm \sqrt{4}\\[1ex]
x+1 & = & \pm 2\\[1ex]
x & = & -1 \pm 2\\[1ex]
x=-1 -2 & \mbox{or} & x=-1+2\\[1ex]
x=-3 & \mbox{or} & x=1
\end{array}
\]
So besides the vertex and the \(y\)-intercept, we found two \(x\)-intercepts: \((-3,0)\) and \((1,0)\).
As extra points to draw a better graph, choose two points on either side of the vertex. We already have \((0,-6)\)
on the right, and we can choose \(x=-2\) on the left: \(f(-2)=2(-2+1)^2-8=2(-1)^2-8=2-8=-6\). So \((-2,-6) \) is another
point on the graph. We summarize the points found in the table below:
\[
\begin{array}{c|c|c|c|c|c}
x &-3 & -2 & -1 & 0 & 1 \\
\hline
f(x) & 0 & -6 & -8 & -6 & 0\\
\hline
& x\mbox{-intercept} & & \mbox{Vertex} & y\mbox{-intercept} &x\mbox{-intercept}
\end{array}
\]
We can now draw an accurate graph:
\[\img{U2_1FA.png}{}{10em}{}\]
\end{example}
\begin{example}
Let \(g(x)=-2(x-2)^2\). We have \(a=-2\), \(h=2\), \(k=0\). So the parabola will open down,
and the vertex is \[V=(2,0).\] Since \(g(0)=-2(0-2)^2=-8\),
the \(y\)-intercept is \((0,-8)\).
To find the \(x\)-intercepts we solve \(-2(x-2)^2=0\), and we find the only
solution \(x=2\).
So there is only one \(x\)-intercept, and it's the same point as the vertex.
This means we have only two points for the graph. To draw a better graph, we select
an \(x\)-value on each side of the vertex. Since the \(x\)-coordinate of the vertex is
\(h=2\), we select \(x=1\) on the left and \(x=3\) on the right.
We find \(g(1)=-2(1-2)^2=-2(-1)^2=-2(1)=-2\)
and \(g(3)=-2(3-2)^2=-2(1)^2=-2(1)=-2\).
So \((1,-2)\) and \((3,-2)\) are points on the graph. We summarize the information:
\[
\begin{array}{c|c|c|c|c}
x &0 & 1 & 2 & 3 \\
\hline
f(x) & -8 & -2 & 0 & -2 \\
\hline
& y\mbox{-intercept} & & \mbox{Vertex and \(x\)-intercept}&
\end{array}
\]
We can now
draw an accurate graph:
\[
\begin{array}{c}
\img{U2_1F8.jpg}{}{10em}{}\\
y=-2(x-2)^2
\end{array}
\]
\end{example}
\begin{example}
Let \(f(x)=(x-1)^2+1\). So \(a=1\), \(h=1\), \(k=1\). The parabola opens up, and the vertex is
\[V=(1,1).\]
Since \(f(0) = (0-1)^2+1=(-1)^2+1=2\), the \(y\) intercept is:
\[y\mbox{-intercept}=(0,2).\]
To find the \(x\) intercepts, we need to solve \((x-1)^2+1=0\). When we try we find:
\[
\begin{array}{rcll}
(x+1)^2+1 & = & 0 \\[1ex]
(x+1)^2 & = & -1\\[1ex]
x+1 & = & \pm \sqrt{-1} & \mbox{not a real number}
\end{array}
\]
So there are no real number solutions, and the graph has no \(x\)-intercepts. As before, we select
two other \(x\)-values on each side of the vertex in order to draw a
better graph. \(f(-1)=(-1-1)^2+1=5\)
and \(f(2)=(2-1)^2+1=2\).
\[
\begin{array}{c}
\img{U2_1F9.jpg}{}{10em}{}\\
y=(x-1)^2+1
\end{array}
\]
\end{example}
Problems
\problem
The vertex of a quadratic function is \(V=(1,-5)\) and the point \((3,2)\) is on the graph.
Find the standard form of the equation.
\begin{sol}
We know that \(h=1\) and \(k=-5\). So the standard form of the
quadratic function is \(f(x)=a(x-1)^2-5\). Now we use the fact that \((3,2)\) is a point
on the graph. We substitute \(x=3\) and \(f(x)=2\), and solve
for \(a\):
\[
\begin{array}{rcll}
2& = & a(2)^2-5\\[1ex]
2 & = & 4a-5\\[1ex]
7 & = & 4a\\[1ex]
a & = & \displaystyle \frac{7}{4}\\
\end{array}
\]
So the standard form of the quadratic function is
\[f(x)=\frac{7}{4}(x-1)^2-5\]
\end{sol}
\mproblem
The vertex of a quadratic function is \(V=(-2,4)\), and the point \((3,-4)\)
is on the graph. Find the standard form of the equation.
\problem
Draw an accurate graph of the quadratic functions, showing the vertex and all intercepts.
\begin{enumerate}
\item
\(f(x)=-(x+2)^2+9\)
\item
\(\displaystyle{g(x)=\frac{1}{2}(x-2)^2}\)
\item
\(p(x)=3x^2-3\)
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item
\(f(x)=-(x+2)^2+9\). The parabola is opening down (because \(a=-1\),
and the vertex is \(V=(-2,9)\). Since
\(f(0)=-(0+2)^2+9=-4+9=5\), the \(y\)-intercept is \((0,5)\). Solving \(f(x)=0\) we find
\[
\begin{array}{rcll}
-(x+2)^2+9 & = & 0 \\[1ex]
9 & = & (x+2)^2\\[1ex]
\pm \sqrt{9} & = & x+2\\[1ex]
-2\pm 3 & = & x\\[1ex]
-2+3 =x & \mbox{or} & -2-3= x\\[1ex]
x=1 & \mbox{or} & x=-5
\end{array}
\]
So the \(x\)-intercepts are \((1,0)\) and \((-5,0)\).
We summarize all the information we have:
\[
\begin{array}{c|c|c|c}
\mbox{Opens} & \mbox{Vertex} & y\mbox{-intercept} & x\mbox{-intercepts}\\
\hline
\mbox{Down} & (-2,9) & (0,5) & (1,0), (-5,0)
\end{array}
\]
Now we can draw an accurate graph:
\[
\begin{array}{c}
\img{U2_1F10.jpg}{}{10em}{}\\
f(x)=-(x+2)^2+9
\end{array}
\]
\item
\(\displaystyle{g(x)=\frac{1}{2}(x-2)^2}\). The parabola is opening up (because \(a=1/2\))
and the vertex is \(V=(2,0)\). So the vertex is also an \(x\)-intercept.
And in fact solving \(\frac{1}{2}(x-2)^2=0\) we find that \(x=2\) is the only solution.
So the graph has only one \(x\)-intercept, that coincides with the vertex.
Since \(g(0)=\frac{1}{2}(0-2)^2=\frac{1}{2}(4)=2\), the
\(y\)-intercept is \((0,2)\). To draw a better graph, we choose two points on either side of the
vertex. Choosing \(x=1\) we get a fraction, \(x=0\) gives us back the \(y\)-intercept,
and \(x=-1\) also gives us a fraction. The first number on
the left side of the vertex that does not give us a fraction
is \(x=-2\), and \(g(-2)=\frac{1}{2}(-2-2)^2=\frac{1}{2}(-4)^2=\frac{1}{2}(16)=8\).
On the right side we choose \(x=4\),
\(g(4)=\frac{1}{2}(4-2)^2=\frac{1}{2}(2)^2=\frac{1}{2}(4)=2\). So \((-2,8)\) and \((4,2)\) are two points on the graph. Here is what we found:
\[
\begin{array}{c|c|c|c|c}
\mbox{Opens} & \mbox{Vertex} & y\mbox{-intercept} & x\mbox{-intercept} & \mbox{Extra points}\\
\hline
\mbox{Up} & (2,0) & (0,2) & (2,0) & (-2,8), (4,2)
\end{array}
\]
We can now draw it:
\[
\begin{array}{c}
\img{U2_1F11.jpg}{}{10em}{}\\
\displaystyle g(x)=\frac{1}{2}(x-2)^2
\end{array}
\]
\item
\(p(x)=-(x+1)^2-1\). Since \(a=-1\), the parabola opens down.
The vertex is \(V=(-1,-1)\). \(p(0)=-(0+1)^2-1=-2\),
so the \(y\)-intercept is \((0,-2)\). Solving \(p(x)=0\) we find
\[
\begin{array}{rcll}
-(x+1)^2-1 & = & 0 \\
-1 & = & (x+1)^2\\
\pm \sqrt{-1} & = & x+1 & \mbox{not a real number}
\end{array}.
\]
So there are no \(x\)-intercepts. We choose \(x=-2\) and \(x=1\) as extra points:
\(p(-2)=-(-2+1)^2-1=-(-1)^2-1=-1-1=-2\), and \(p(1)=-(1+1)^2-1=-(2)^2-1=-4-1=-5\).
So \((-2,-2)\) and \((1,-5)\) are on the graph.
\[
\begin{array}{c|c|c|c|c}
\mbox{Opens} & \mbox{Vertex} & y\mbox{-intercept} & x\mbox{-intercepts} & \mbox{Extra points}\\
\hline
\mbox{Down} & (-1,-1) & (0,-2) & \mbox{None} & (-2,-2), (1,-5)
\end{array}
\]
\[
\begin{array}{c}
\img{U2_1F12.jpg}{}{10em}{}\\
p(x)=-(x+1)^2-1
\end{array}
\]
\end{enumerate}
\end{sol} \mproblem
Draw an accurate of the quadratic functions, showing the vertex and all intercepts.
\begin{enumerate}
\item
\(f(x)=-(x+3)^2+1\)
\item
\(g(x)=2(x-1)^2+1\)
\item
\(p(x)=2(x+1)^2\)
\end{enumerate}