Two Important Graphs Revision Notes for VCE SSCE Mathematical Methods

Two Important Graphs

The graph of $y^2 = x$

Understanding the basic graph

The equation $y^2 = x$ creates an interesting type of parabola that opens horizontally to the right, rather than vertically like the more familiar $y = x^2$ . This is because $y$ is squared rather than $x$ .

To understand this graph, let's examine some coordinate pairs. When we choose values for $y$ between $-4$ and $4$ , we can calculate the corresponding $x$ -values:

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Notice that both positive and negative $y$ -values give the same $x$ -value. For example, when $y = 2$ or $y = -2$ , we get $x = 4$ . This symmetry is a key feature of this graph and occurs because squaring any number (positive or negative) gives a positive result.

When we plot these points and join them with a smooth curve, we create the following parabola:

Key features of $y^2 = x$

This graph has several important characteristics:

Shape: It forms a parabola opening to the right
Vertex: The turning point is located at the origin $(0, 0)$
Axis of symmetry: The $x$ -axis serves as the line of symmetry
Relationship to $y = x^2$ : This graph is the reflection of $y = x^2$ across the line $y = x$

The parabola has two branches - one extending upward into the positive $y$ region and one extending downward into the negative $y$ region. Both branches curve away from the $y$ -axis as $x$ increases.

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Unlike the familiar vertical parabola $y = x^2$ , this horizontal parabola $y^2 = x$ has its axis of symmetry along the $x$ -axis. This is a fundamental difference that affects how we work with transformations and find intercepts.

Transformations of $y^2 = x$

Just like other functions, we can transform the basic parabola $y^2 = x$ by applying translations and dilations. The general form is:

(y - k)^2 = a^2(x - h)

where:

The vertex is located at the point $(h, k)$
The axis of symmetry is the horizontal line $y = k$
The value of $a$ affects the width of the parabola

Worked example: Translating the parabola

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Worked Example: Sketching $(y - 4)^2 = x + 3$

Let's sketch the graph of $(y - 4)^2 = x + 3$ .

Solution:

This graph is formed by taking the basic parabola $y^2 = x$ and translating it $3$ units to the left and $4$ units upward.

The vertex has coordinates $(-3, 4)$ .

To find the $y$ -intercepts (where $x = 0$ ):

(y - 4)^2 = 3

y - 4 = \pm\sqrt{3}

y = 4 \pm \sqrt{3}

To find the $x$ -intercept (where $y = 0$ ):

16 = x + 3

x = 13

The graph shows the parabola with vertex at $(-3, 4)$ , passing through the $y$ -axis at $4 + \sqrt{3}$ and $4 - \sqrt{3}$ , and crossing the $x$ -axis at $13$ .

Worked example: Completing the square

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Worked Example: Rearranging $y^2 + 2y = 2x + 3$

Let's sketch the graph of $y^2 + 2y = 2x + 3$ .

Solution:

First, we need to rearrange this equation into the standard form by completing the square:

y^2 + 2y = 2x + 3

y^2 + 2y + 1 = 2x + 4

(y + 1)^2 = 2(x + 2)

This graph is obtained from $y^2 = x$ by:

Dilating by a factor of $\sqrt{2}$ from the $x$ -axis
Translating $2$ units to the left
Translating $1$ unit downward

The vertex has coordinates $(-2, -1)$ .

To find the $y$ -intercepts (where $x = 0$ ):

(y + 1)^2 = 4

y + 1 = \pm 2

y = -1 \pm 2

So $y = 1$ or $y = -3$

To find the $x$ -intercept (where $y = 0$ ):

1 = 2x + 4

x = -\frac{3}{2}

The graph of $y = \sqrt{x}$

Understanding the basic graph

The function $y = \sqrt{x}$ can also be written as $y = x^{\frac{1}{2}}$ . This graph represents the upper half of the parabola $y^2 = x$ .

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Domain Restriction: The square root function is only defined for non-negative values, so $x \geq 0$ . You cannot take the square root of a negative number in the real number system.

The graph starts at the origin $(0, 0)$ and curves upward to the right. Some key points include $(1, 1)$ and $(4, 2)$ .

The general form

All graphs of the form:

y = a\sqrt{x - h} + k

have the same basic shape as $y = \sqrt{x}$ , but with transformations applied:

$h$ shifts the graph horizontally (the starting point moves to $x = h$ )
$k$ shifts the graph vertically
$a$ dilates the graph vertically and can reflect it if negative

The endpoint of the transformed graph is at $(h, k)$ .

Domain and range

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Understanding Domain and Range

When working with square root functions, always consider:

Domain: The set of $x$ -values for which the function is defined (what values can go under the square root)
Range: The set of possible $y$ -values the function can produce

For $y = \sqrt{x}$ , the domain is $x \geq 0$ and the range is $y \geq 0$ .

Worked example: Dilation and translation

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Worked Example: Sketching $y = 3\sqrt{x + 1} - 6$

Let's sketch the graph of $y = 3\sqrt{x + 1} - 6$ .

Solution:

When $x = 0$ :

y = -3

When $y = 0$ :

3\sqrt{x + 1} - 6 = 0

3\sqrt{x + 1} = 6

\sqrt{x + 1} = 2

x + 1 = 4

x = 3

Explanation:

This graph is formed by:

Dilating the graph of $y = \sqrt{x}$ from the $x$ -axis by a factor of $3$
Translating $1$ unit to the left
Translating $6$ units downward

The function is defined for $x \geq -1$ (since we need $x + 1 \geq 0$ ).

The range is all values greater than or equal to $-6$ , written as $y \geq -6$ .

Worked example: Reflection in the $x$ -axis

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Worked Example: Sketching $y = -2\sqrt{x - 1} + 3$

Let's sketch the graph of $y = -2\sqrt{x - 1} + 3$ .

Solution:

When $y = 0$ :

-2\sqrt{x - 1} + 3 = 0

2\sqrt{x - 1} = 3

Squaring both sides:

4(x - 1) = 9

x = \frac{9}{4} + 1 = \frac{13}{4}

Explanation:

This graph is formed by:

Dilating the graph of $y = \sqrt{x}$ from the $x$ -axis by a factor of $2$
Reflecting in the $x$ -axis (due to the negative sign)
Translating $1$ unit to the right
Translating $3$ units upward

The function is defined for $x \geq 1$ .

The range is all values less than or equal to $3$ , written as $y \leq 3$ . Notice how the reflection changes the range - the graph now extends downward rather than upward.

The graph of $y = \sqrt{-x}$

An important variation is the function:

y = \sqrt{-x} \text{ for } x \leq 0

This creates a graph that is the reflection of $y = \sqrt{x}$ in the $y$ -axis.

The graph extends to the left from the origin, curving upward into the second quadrant.

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All graphs of the form:

y = a\sqrt{-(x - h)} + k

follow the same basic shape as $y = \sqrt{-x}$ . The negative sign inside the square root causes the reflection in the $y$ -axis.

Worked example: Graph of $y = \sqrt{-x}$ transformed

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Worked Example: Sketching $y = \sqrt{2 - x} + 3$

Let's sketch the graph of $y = \sqrt{2 - x} + 3$ .

Note: We can rewrite this as $y = \sqrt{-(x - 2)} + 3$

Solution:

When $x = 0$ :

y = \sqrt{2} + 3

Explanation:

We can express the rule as:

y = \sqrt{-(x - 2)} + 3

This function is defined for $x \leq 2$ (since we need $2 - x \geq 0$ ).

The range is all values greater than or equal to $3$ , written as $y \geq 3$ .

The graph starts at the point $(2, 3)$ and extends to the left, with the curve rising as $x$ decreases.

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Key Points to Remember:

The graph of $y^2 = x$ is a horizontal parabola opening to the right with vertex at the origin and axis of symmetry along the $x$ -axis
The general form $(y - k)^2 = a^2(x - h)$ has its vertex at $(h, k)$ and axis of symmetry at $y = k$
The graph of $y = \sqrt{x}$ is the upper half of the parabola $y^2 = x$ and is defined only for $x \geq 0$
For square root functions, always identify the domain (valid $x$ -values) and range (possible $y$ -values)
The graph of $y = \sqrt{-x}$ is the reflection of $y = \sqrt{x}$ in the $y$ -axis and is defined for $x \leq 0$

Two Important Graphs (VCE SSCE Mathematical Methods): Revision Notes