3.02 Graphs of parabolas

When we think of the word parabola, we usually associate it with a function we're very familiar with, the quadratic function.

Quadratic function recap

Quadratic functions can be written in each of the following forms:

• $y=ax^2+bx+c$y=ax2+bx+c, which is known as the general form
• $y=a\left(x+\alpha\right)\left(x+\beta\right)$y=a(x+α)(x+β), which is known as the factored form
• $y=a\left(x-h\right)^2+k$y=a(x−h)2+k, which is known as the turning point form where the turning point is ($h$h, $k$k) 

 

The graph of the relation $y^2=x$y2=x is also a parabola

Consider the relation $y^2=x$y2=x. This graph is not a function. 

Firstly, why is this called a relation, and not a function?

Let's look at a few points on the graph.

• If $y^2=1$y2=1, then $y$y can take on the values of $-1$−1 and $1$1. This means that for $x=1$x=1 there are two $y$y value outputs, rather than just one.
• If $y^2=4$y2=4, then y can take on the values of $-2$−2 and $2$2. This means that for $x=4$x=4 there are two $y$y value outputs, rather than just one.

In fact, this will be true for all values of $x>0$x>0 that we try. So since there are two outputs for every input, $y^2=x$y2=x can't be considered a function.

 

The graph of $y^2=x$y2=x

When you look at the graph below, it should remind you of another, familiar graph.

This looks like the graph of a quadratic function that has been rotated to lie on its side.

In fact, $y^2=x$y2=x is a reflection of $y=x^2$y=x2 across the line $y=x$y=x.

 

Can we restrict $y^2=x$y2=x to make a function?

Let's rearrange $y^2=x$y2=x so that $y$y is the subject:

$y=\pm\sqrt{x}$y=±√x

We can break this up into two separate functions: $y=\sqrt{x}$y=√x and $y=-\sqrt{x}$y=−√x

If we sketch the graphs of each of these separately, we can see that each graph has one output for each input.

The graph of $y=\sqrt{x}$y=√x

The graph of $y=-\sqrt{x}$y=−√x

 

Practice question

Question 1

 

Translation of parabolas

We could apply translations to the graph of $x=y^2$x=y2, just as we did in quadratic functions and form an equation analogous to turning point form: $x=a\left(y-k\right)^2+h$x=a(y−k)2+h

Let's look at the orientation possibilities of parabolas that can be constructed using the forms:

$y=a\left(x-h\right)^2+k$y=a(x−h)2+k and $x=a\left(y-k\right)^2+h$x=a(y−k)2+h

Examples of these are shown below. Look carefully at the the coordinates of each vertex and how that matches up with the corresponding equation.

For instance, in diagram (B) we have a vertical parabola and there is a horizontal translation of $2$2 units to the left and $5$5 units up. There is also a reflection to deal with - the parabola is opening downward, which means the value of $a$a is negative. This matches the form of the corresponding equation, $y=-\left(x+2\right)^2+5$y=−(x+2)2+5 .

In diagram (C) there is a left opening horizontal parabola with the vertex at ($-5$−5, $2$2). Left opening means the value of $a$a is negative and the equation will be $x=-\left(y-2\right)^2-5$x=−(y−2)2−5 . 

Summary:

• Each graph has its vertex at $\left(h,k\right)$(h,k)
• If the graph is of the form $y=...$y=..., then the axis of symmetry is vertical with equation $x=h$x=h
• If the graph is of the form $x=...$x=..., then the axis of symmetry is horizontal with equation $y=k$y=k
• If $a>0$a>0 the graph opens upwards if it's $y=...$y=... or to the right if it's $x=...$x=...
• If $a<0$a<0 the graph opens downward if it's $y=...$y=...or to the left if it's $x=...$x=...

Some examples:

Equation	Vertex	Opens	Think
$y=\left(x-3\right)^2+2$y=(x−3)2+2	($3$3, $2$2)	upwards	The $-3$−3 inside the bracket translates the $x$x values $3$3 right. The $+2$+2 translates the $y$y values $2$2 up
$y=-\left(x+2\right)^2-4$y=−(x+2)2−4	($-2$−2, $-4$−4)	downwards	The $+2$+2 inside the bracket translates the $x$x values $2$2 left The $-4$−4 translates the $y$y values 4 down
$x=\left(y-4\right)^2+5$x=(y−4)2+5	($5$5, $4$4)	right	The $-4$−4 inside the bracket translates the $y$y values 4  up. The $+5$+5 translates the $x$x values $5$5 right
$x=-\left(y+3\right)^2-2$x=−(y+3)2−2	($-2$−2, $-3$−3)	left	The $+3$+3 inside the bracket translates the $y$y values $3$3 down. The $-2$−2 translates the $x$x values $2$2 left

Worked example

Example 1

(a) Determine the equation of a horizontal parabola that has its vertex at $\left(-3,4\right)$(−3,4), and its arms opening to the left.

Think: Horizontal and arms opening left means the equation is in the form $x=...$x=... and $a$a is negative. The vertex of $\left(-3,4\right)$(−3,4) means the $x$x values of base equation $x=y^2$x=y2 are translated $3$3 left and the $y$y values are translated $4$4 up. 

Do: Write: the equation is  $x=-\left(y-4\right)^2-3$x=−(y−4)2−3 .

Reflect: This is not the only possible choice - any equation of the form $x=-a\left(y-4\right)^2-3$x=−a(y−4)2−3 (where $a$a is a positive number) will also have those attributes. The coefficient $a$a is a dilation factor and will dilate the graph by a factor of $a$a from the $y$y-axis (as $a$a increases the graph gets skinnier because the $x$x values are being multiplied by $a$a - check it by graphing some functions on your calculator!).

 

(b) State the domain and range of the equation from part (a).

Think: The vertex is at $\left(-3,4\right)$(−3,4) and the curve opens to the left. This means $x$x values are left of $-3$−3 and all $y$y values are in the range. 

Do: Write: the domain is  $x:x\in\mathbb{R},x\le-3$x:x∈ℝ,x≤−3 and the range is $y:y\in\mathbb{R}$y:y∈ℝ.

Reflect: The domain could be written as $\left(-\infty,-3\right]$(−∞,−3] and the range could be written $\left(-\infty,\infty\right)$(−∞,∞) in interval notation. 

 

Different forms

Just as with our quadratic functions we can be asked to graph parabolas from general form: $y=ax^2+bx+c$y=ax2+bx+c or $x=ay^2+by+c$x=ay2+by+c. Recall we can change from general form to turning point form by completing the square. We can also find the turning point from this form by first finding the equation of the line of symmetry, using the formula $x=\frac{-b}{2a}$x=−b2a​ for vertical parabolas or $y=\frac{-b}{2a}$y=−b2a​ for horizontal parabolas. Then we can substitute this back into the equation to find the turning point.

 

Practice questions

Question 2

Question 3

Question 4

Question 5

Question 6