Hong Kong
Stage 4 - Stage 5

# Graphing Parabolas (based on vertex and transformations)

Lesson

## The four basic directions

The orientation possibilities of the parabola, with axes parallel to the coordinate axes, can be constructed with the forms  $y-k=\pm\left(x-h\right)^2$yk=±(xh)2 and $x-k=\pm\left(y-h\right)^2$xk=±(yh)2

Examples of the 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), there is a horizontal translation of $2$2 units to the left and $5$5 units up. There is also an inversion to deal with - the parabola is opening downward and the function becomes more negative as the values of $x$x move away from the line of symmetry $x=-2$x=2. Consequently, corresponding equation becomes $y-5=-\left(x+2\right)^2$y5=(x+2)2.

So if we wanted to construct an equation of a parabola that had its vertex at $\left(-3,4\right)$(3,4), and its arms opening to the left, we could choose $x+3$x+3=$-\left(y-4\right)^2$(y4)2. There are others of course - for example the equation given by $x+3=-k\left(y-4\right)^2$x+3=k(y4)2, where $k$k is any positive number will also have those attributes.

## Understanding the concept

Without the parabola's equation being in one of the above forms, we might think that we cannot obtain any information about it. This is really not true.

In fact we can immediately know which way the parabola is opening when the equation is given in one of the two basic forms of $y=ax^2+bx+c$y=ax2+bx+c or $x=ay^2+by+c$x=ay2+by+c.

Let's think about the parabola $y=x^2+6x+16$y=x2+6x+16.

When $x$x gets large, $x^2$x2 gets larger at a rate faster than the linear term $6x$6x and, of course, the unchanging constant term $16$16. This is the key concept to get hold of. As the curve grows out from the vertex, the rapidly increasing size of the function value is mostly explained by the squared term. This means that the parabola must be opening upwards.

Alternatively, if a negative sign was attached to the squared term, so that $y=-x^2+6x+16$y=x2+6x+16, then the function values must eventually become negative as the curve moves out from the vertex. This means the curve is opening downwards.

The big lesson is this - Where the curve moves to, after leaving the vertex, is principally determined by the term in the quadratic with the highest power.

As a useful exercise, we could reorganise the equation $y=x^2+6x+16$y=x2+6x+16:

 $y$y $=$= $x^2+6x+16$x2+6x+16 $=$= $x^2+6x+9+7$x2+6x+9+7 $=$= $\left(x+3\right)^2+7$(x+3)2+7 $\therefore$∴     $y-7$y−7 $=$= $\left(x+3\right)^2$(x+3)2

The parabola in vertex form is shown as type (a) above, with a vertex of $\left(-3,7\right)$(3,7) and opening upwards.

We could do the same with the parabola given by $y=-x^2+6x+16$y=x2+6x+16:

 $y$y $=$= $-x^2+6x+16$−x2+6x+16 $=$= $-\left(x^2-6x-16\right)$−(x2−6x−16) $=$= $-\left(x^2-6x+9-25\right)$−(x2−6x+9−25) $=$= $-\left[\left(x-3\right)^2-25\right]$−[(x−3)2−25] $=$= $-\left(x-3\right)^2+25$−(x−3)2+25 $\therefore$∴    $y-25$y−25 $=$= $-\left(x-3\right)^2$−(x−3)2

Again we can see it's a parabola of variety (b) above, with vertex $\left(3,25\right)$(3,25) and opening downwards.

The same is true for $x=ay^2+by+c$x=ay2+by+c.

If $a$a is positive, the curve will open to the right principally because the term $ay^2$ay2 causes the $x$x values to eventually become positive.  If $a$a is negative, the opposite happens - the curve will open to the left.

A parabola will always open up around its focus, no matter which way its orientated.

The general parabola given by $\left(x-h\right)^2=4a\left(y-k\right)$(xh)2=4a(yk), shows a as the focal length, and this means that the focus has coordinates $\left(h,k+a\right)$(h,k+a).

The same idea applies to the other varieties. Here are some examples.

##### Example 1

The parabola $y-5$y5=$2\left(x-3\right)^2$2(x3)2 can be re-expressed as $\left(x-3\right)^2=\frac{1}{2}\left(y-5\right)$(x3)2=12(y5). The focal length $a=\frac{1}{8}$a=18 and the coordinates of the focus are $\left(3\frac{1}{8},5\right)$(318,5)

##### Example 2

The parabola $x-1=-\frac{1}{24}\left(y+7\right)^2$x1=124(y+7)2, which is a variety (d) above, can be re-expressed as  $\left(y+7\right)^2=-24\left(x-1\right)$(y+7)2=24(x1). This means that it is left opening, with a focal length $a=6$a=6 and a vertex at $\left(-5,-7\right)$(5,7).

#### More Examples

##### Question 1

Consider the parabola represented by the equation $y-4=\left(x+5\right)^2$y4=(x+5)2.

1. What are the coordinates of the vertex?

Give your answer in the form $\left(a,b\right)$(a,b).

2. In which direction does this parabola open?

To the right

A

Downwards

B

Upwards

C

To the left

D

##### Question 2

Consider the parabola represented by the equation $x-4=\left(y+2\right)^2$x4=(y+2)2.

1. What are the coordinates of the vertex?

Give your answer in the form $\left(a,b\right)$(a,b).

2. In which direction does this parabola open?

Upwards

A

To the left

B

Downwards

C

To the right

D

##### Question 3

1. In which direction does the parabola represented by the equation $y=4x^2+3x+5$y=4x2+3x+5 open?

down

A

up

B

right

C

left

D
2. In which direction does the parabola represented by the equation $y=-3x^2+4x-5$y=3x2+4x5 open?

left

A

down

B

right

C

up

D
3. In which direction does the parabola represented by the equation $x=2y^2-9y+5$x=2y29y+5 open?

left

A

down

B

right

C

up

D
4. In which direction does the parabola represented by the equation $x=-2y^2-3y+5$x=2y23y+5 open?

up

A

right

B

down

C

left

D