The graph of quadratic functions

The graph of a quadratic function

f(x)=ax^2+bx+c

is shaped like a $\cup$ or $\cap$ , and is called parabola. The highest point of the $\cap$ -graph, or the lowest point of the $\cup$ -graph, is called the vertex of $f$ . The vertices for the two parabolas shown below are indicated by a small dot.

The simplest quadratic function is

f(x)=x^2

We call it the reference quadratic function. Using a table of values to draw its graph

\begin{array}{r|l} x & y=x^2 \\ \hline -3 & 9\\ -2 & 4\\ -1 & 1\\ 0 & 0\\ 1 & 1\\ 2 & 4\\ 3 & 9\\ \end{array}

we get the figure below. It has a $\cup$ shape and the vertex is at the origin: $S(0|0)$ . Furthermore, the graph passes through the points $P(-1|1)$ and $Q(1|1)$ .

Determining the graph of a quadratic function in standard form

f(x)=ax^2+bx+c

is not straight forward, as not all coefficients $a, b$ and $c$ have a clear geometric interpretation. For the parameters of the vertex form

f(x)=A(x-v)^2+b

however, there is a clear geometric interpretation, and is therefore the preferred form if we need to draw the graph of $f$ . We have the following interpretation of the parameters $A, v$ , and $b$ :

$x^2 \rightarrow Ax^2$ : Stretch the graph of the reference function $x^2$ in $y$ -direction by a factor of $A$ . If $A<0$ , the graph of $x^2$ is stretched and also reflected about the $x$ -axis. In particular, the vertex of the new graph is still at the origin: $S(0\vert 0)$ .
$Ax^2 \rightarrow A(x-v)^2$ : Shift the graph we obtained above to the right ( $v>0$ ) or to the left ( $v<0$ ) by $v$ .The vertex of the new graph is now at $S(v\vert 0)$ .
$A(x-v)^2 \rightarrow A(x-v)^2+b$ : Shift the graph obtained above up ( $b>0$ ) or down ( $b<0$ ) by $b$ .The vertex of the new graph is now at $S(v\vert b)$ .

So you see, the parameters $v$ and $b$ are simply the coordinates of the vertex $S$ of the parabola.

Let's try to understand why these geometric interpretations of the parameters $A, v$ and $b$ are correct. Below you see the quadratic function in vertex form represented as a machine.

Notice that $A$ multiplies the output of the $(\,)^2$ -machine, and $b$ adds to it. Because the output of a machine is the $y$ -coordinate of its graph, we see that $A$ stretches every point of the graph of $x^2$ by the factor $A$ in $y$ -direction, and $b$ shifts the points vertically up or down. So the effect of $A$ and $b$ on the reference graph $x^2$ should be clear.

To understand the effect of $v$ , let's find the position of the vertex of $f(x)=A(x-v)^2$ , that is, $b=0$ . So we know from above, that the vertex will be on the $x$ -axis. So let's calculate the $x$ -intercept of $f$ :

\begin{array}{lll} A(x-v)^2=0 \quad | :A, \sqrt{\phantom{x}}\\ x-v = 0\quad | +v\\ x=v \end{array}

and we see that the vertex of $f$ is at $S(v|0)$ . In other words, for $v>0$ the reference graph is shifted to the right, and for $v<0$ the reference graph is shifted to the left.

Let us summarise and formulate the theorem

Theorem 1

Consider the vertex form of a quadratic function

f(x)=A(x-v)^2+b

The graph of $f$ is a stretched version of the reference graph $x^2$ (stretched in $y$ -direction by the factor of $A$ ). Its vertex is at $S(v|b)$ . The graph of $f$ is $\cup$ -shaped for $A>0$ and $\cap$ -shaped for $A<0$ .

Example 1

Consider the function $f(x)=2x^2+4x-4$ .

Determine the stretching factor in $y$ -direction and the coordinates of its vertex.
Sketch the graph without a table of values, just by considering the succession of transformations applied to $x^2$ .

Solution

We bring $f$ into vertex form:

\begin{array}{lll} f(x) & = & 2x^2+4x-4\\ & = & 2(x^2+2x)-4\\ & = & 2((x+1)^2-1^2)-4\\ & = & 2(x+1)^2-6\\ & = & 2(x-(-1))^2-6\\ \end{array}

Thus, $f$ is stretched in $y$ -direction by $A=2$ , and the vertex is at $S(-1|-6)$ ( $v=-1, b=-6$ ).
First, draw the graph of the reference function, and stretch it by a factor of $2$ in $y$ -direction. Then shift this new graph to the point $S(-1|-6)$ . See figure below.

Exercise 1: Function equation \rightarrow graph

Solution

Let $r$ denote the reference function: $r(x)=x^2$ , and $S$ the vertex.

$f(x)=(x-2)^2+3$ , thus $A=1, v=2, b=3$ , and therefore $\underline{S(2|3)}$ .That is, graph $r$ is shifted to $S(2|3)$ .
$g(x)=-2(x+2)^2-1$ , thus $A=-2, v=-2, b=-1$ , and therefore $\underline{S(-2|-1)}$ .That is, graph $r$ is stretched in $y$ -direction by the factor of $2$ , reflected about the $x$ -axis, and shifted to $S(-2|-1)$ .
$h(x)=x^2-2x+5 = (x-1)^2+4$ , thus $A=1, v=1, b=4$ and therefore $\underline{S(1|4)}$ That is, graph $r$ is shifted to $S(1|4)$ .
$i(x)=(x+2)^2$ , thus $A=1, v=-2, b=0$ , and therefore $\underline{S(-2|0)}$ .That is, graph $r$ shifted to $S(-2|0)$ .
$k(x)=-0.5x^2+1 = -0.5(x-0)^2+1$ , thus $A=-0.5, v=0, b=1$ and therefore $\underline{S(0|1)}$ .That is, graph $r$ is stretched in $y$ -direction by the factor of $0.5$ , reflected about the $x$ -axis, and shifted to $S(0|1)$ .
$j(x)=\frac{3}{4}x^2-6x+\frac{15}{2} =\frac{3}{4}(x-4)^2-4.5$ , thus $A=\frac{3}{4}, v=4, b=-4.5$ and therefore $\underline{S(4|-4.5)}$ .That is, graph $r$ is stretched in $y$ -direction by the factor of $3/4$ , and shifted to $S(4|-4.5)$ .
$l(x)=3(2x-4)^2+2=3(2(x-2))^2+2=3\cdot 2^2(x-2)^2+2=12(x-2)^2+2$ , thus $A=12, v=2, b=2$ and therefore $\underline{S(2|2)}$ . That is, graph $r$ is stretched in $y$ -direction by the factor of $12$ , and shifted to $S(2|2)$ .

Determine the coordinates of the vertex and the stretching factor in $y$ -direction, then use this information to sketch the graph:

$f(x)=(x-2)^2+3$
$g(x)=-2(x+2)^2-1$
$h(x)=x^2-2x+5$
$i(x)=(x+2)^2$
$k(x)=-0.5x^2+1$
$j(x)=\frac{3}{4}x^2-6x+\frac{15}{2}$
$l(x)=3(2x-4)^2+2$

Exercise 2: Graph \rightarrow function equation

Solution

$f$ : Diagram $\rightarrow S(1|0)$ , thus $f(x)=A(x-1)^2+0$ . To find $A$ , we see that $f(2)=1 \rightarrow A(2-1)^2+0=1 \rightarrow A=1$ . Thus, we have $f(x)=1\cdot (x-1)^2+0=\underline{(x-1)^2}$ .The value of $A$ can also be directly inferred from the diagram if you compare with the reference function $x^2$ (one to the right from the vertex, one up).
$g$ : Diagram $\rightarrow S(-3|3)$ , thus $g(x)=A(x+3)^2+3$ . To find $A$ , we see that $g(-2)=2$ , thus $A(-2+3)^2+3=2 \rightarrow A=-1$ . Thus we have $g(x)=\underline{-(x+3)^2+3}$ . The value $A=-1$ can also be derived directly by comparing with the reference function $x^2$ .
$h$ : Diagram $\rightarrow S(0|0)$ , thus $h(x)=A(x-0)^2+0=Ax^2$ . To find $A$ , we see that $h(2)=1$ and thus $A\cdot 2^2=1\rightarrow A=\frac{1}{4}$ . Thus we have $h(x)=\frac{1}{4}x^2$ . Again, by comparing with the reference function $x^2$ , we could infer $A$ directly (two to the right from the vertex, 4 up, so we have to scale by $1/4$ ).
$k$ : Diagram $\rightarrow S(5|-3)$ thus $k(x)=A(x-5)^2-3$ . To find $A$ , we see that $k(6)=0$ and thus $A(6-5)^2-3=0 \rightarrow A=3$ . Thus we have $k(x)=\underline{3(x-5)^2-3}$ . Again, we can infer $A$ directly by comparing with the reference function.

Determine the vertex form of the graphs shown below by first identifying the coordinates of the vertex on the diagram, and then finding the scaling factor in $y$ -direction.

Exercise 3

The graph of the function $f(x)=x^2$ is stretched in $y$ -direction by a factor of $10$ , reflected about the $x$ -axis, shifted $4$ to the right and $5$ upwards. Determine the function equation of the new graph.

Solution

$A=-10, v=4, b=5$ , thus $f(x)=\underline{-10(x-4)^2+5}$

Exercise 4

The $x$ - and $y$ -intercepts of the parabola $f$ .
The point of intersections between the parabola $f$ and the straight line $g$ .
The points of intersection between the parabolas $f$ and $h$ .

Solution

We first determine the function equations.

graph $g$ has slope $\Delta y / \Delta x = 1/2=0.5$ and $y$ -intercept $1$ , thus it is $g(x)=0.5x+1$
For graph $f$ it is $A=1, v=2, b=-2$ , thus $f(x)=(x-2)^2-2$
For graph $h$ it is $A=-1, v=1, b=0$ , thus $h(x)=-(x-1)^2$

$y$ -intercept of $f$ is $f(0)=(0-2)^2-2=\underline{2}$ (see diagram). For the $x$ -intercept, find $x$ with $f(x)=(x-2)^2-2=0$ , thus $(x-2)^2=2$ and therefore $x_1=2+\sqrt{2}=\underline{3.41}$ and $x_2=2-\sqrt{2}=\underline{0.59}$ .
The points of intersections between $g$ and $f$ . Find $x$ with $g(x)=f(x)$ : $\begin{array}{rll} 0.5x+1 & = & (x-2)^2-2\\ 0.5x+1 & = & x^2-4x+2\\ 0 & = & x^2-4.5x+1\\ \end{array}$ Midnight formula: $x_1=4.27$ and $x_2=0.23$ . The corresponding $y$ -coordinates are $y_1=g(4.27)=3.14$ and $y_2=f(0.23)=1.12$ . Thus, the points of intersection are $\underline{A_1(0.23|1.12)}$ and $\underline{A_2(4.27|3.14)}$ .
The points of intersections between $f$ and $h$ : Find $x$ with $h(x)=f(x)$ : $\begin{array}{rll} -(x-1)^2 & = &(x-2)^2-2\\ -x^2+2x-1 & = & x^2-4x+2\\ 0 & = & 2x^2-6x+3\\ \end{array}$ Midnight formula: $x_1=2.37$ and $x_2=0.63$ . The corresponding $y$ -coordinates are $y_1=h(2.37)=-1.88$ and $y_2=h(0.63)=-0.14$ . Thus, the points of intersection are $\underline{C_1(0.63|-0.14)}$ and $\underline{C_2(2.37|-1.88)}$ .

Exercise 5

The graph of the function $f(x)=ax^2+bx+c$ has the vertex $S$ and passes through the point $P$ . Determine the parameters $a, b$ and $c$ of the standard form:

$S(-1|-5)$ , $P(3|11)$
$S(4|12)$ , $P(0|4)$

Solution

We start with the vertex form.

$S(-1|-5)$ , thus $v=-1, b=-5$ and therefore $f(x)=A(x+1)^2-5$ . To find $A$ use $f(3)=11$ (as $P(3|11)$ is on the graph of $f$ ), thus $A(3+1)^2-5=16A-5=11$ and therefore $A=1$ . It follows $f(x)=(x+1)^2-5=x^2+2x-4$ and therefore $\underline{a=1, b=2, c=-4}$
$S(4|12)$ , thus $v=4, b=12$ and therefore $f(x)=A(x-4)^2+12$ . With $f(0)=4$ follows $A(0-4)^2+12=16A+12=4$ , thus $A=-0.5$ . We have $f(x)=-0.5(x-4)^2+12=-0.5x^2+4x+4$ , thus $\underline{a=-0.5, b=4, c=4}$

Exercise 6

Find the parameter $s$ of the function $f(x)=x^2-sx+4$ such that the vertex of the graph of $f$ has the $y$ coordinate $2$ . What is the corresponding $x$ -coordinate?

Solution

The vertex form of $f(x)=x^2-sx+4$ is $f(x)=\left(x-\frac{s}{2}\right)^2-\left(\frac{s}{2}\right)^2+4$ . So the vertex $S(v\vert B)$ has $v=\frac{s}{2}$ and $b=4-\left(\frac{s}{2}\right)^2=4-\frac{s^2}{4}$ . Since the $y$ -coordinate of the vertex is $2$ , we need to find $s$ such that $b=4-\frac{s^2}{4}=2 \rightarrow s=\sqrt{8}$ . So it is $f(x)=\underline{x^2-\sqrt{8}x +4}$ and the $x$ coordinate of the vertex is $v=\frac{s}{2}=\frac{\sqrt{8}}{2}=\underline{1.41...}$ .

Exercise 7

The product of two numbers with difference $1$ should be as small as possible. Find the two numbers. Hint: vertex of quadratic function

Solution

Let $x$ be the smaller number, so the other number is $x+1$ . The product of the two numbers is thus $f(x)=x(x+1)$ , and we need to find $x$ such that $f$ is as small as possible. $f$ is a quadratic function with vertex form

\begin{array}{lll} f(x) &= & x(x+1)\\ &= & (x+0. 5)^2-0.25\\ \end{array}

Since $A=1$ , the vertex $S(-0.5\vert -0.25)$ is the lowest point, and thus the $x$ -value we are looking for is $x=-0.5$ . The two numbers are therefore $\underline{-0.5}$ and $\underline{0.5}$ .

Exercise 8

Which rectangle with a perimeter of $14cm$ has the largest area? Hint: vertex of quadratic function

Solution

Let $x$ and $y$ be the side lengths of the rectangle. Thus

2x+2y=14

and

y=7-x

The area is thus

f(x)=xy = x(7-x)

which is a quadratic function with vertex form

\begin{array}{lll} f(x) &= &x(7-x)\\ & =& -x^2+7x\\ & =& -(x^2-7x) \\ &= & -(x-3. 5)^2+12.25\\ \end{array}

Since $A=-1$ , the vertex $S(3.5\vert 12.25)$ is the highest point, and so the area is largest for $x=\underline{3.5}$ . The other side has length $7-3.5=\underline{3.5}$ , therefore, it is a square.