Graphing Quadratics Revision Notes for VCE SSCE Mathematical Methods

Graphing Quadratics

Introduction to quadratic functions

A quadratic function is a mathematical relationship where the highest power of the variable is 2. Quadratics can be written in two main forms:

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Polynomial form: $y = ax^2 + bx + c$

where $a$ , $b$ and $c$ are constants and $a \neq 0$ .

Turning point form: $y = a(x - h)^2 + k$

Both forms describe the same type of graph, called a parabola. The turning point form is particularly useful for graphing because it immediately tells us where the vertex is located and how the graph has been transformed from the basic parabola $y = x^2$ .

The basic parabola $y = x^2$

The simplest quadratic function is $y = x^2$ . Let's look at its key features by first creating a table of values.

When we plot these points and connect them with a smooth curve, we get a U-shaped graph called a parabola.

Key features of $y = x^2$

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The graph of $y = x^2$ has several important characteristics:

Shape: The graph is called a parabola and has a smooth, curved U-shape.
Vertex (turning point): The lowest point on the graph is at the origin $(0, 0)$ . This is where the parabola changes direction.
Axis of symmetry: The parabola is perfectly symmetrical about the $y$ -axis (the line $x = 0$ ). If you folded the graph along this line, both halves would match exactly.
Range: All $y$ -values are positive or zero. The minimum value of $y$ is $0$ , occurring at the vertex.
Direction: The parabola opens upward.

Transformations of $y = x^2$

We can transform the basic parabola $y = x^2$ by changing the values in the equation. These transformations help us graph any quadratic function.

Effect of the coefficient $a$

When we change the coefficient $a$ in $y = ax^2$ , we affect the width and direction of the parabola.

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When $a > 1$ : The parabola becomes narrower (steeper) than $y = x^2$ . For example, $y = 2x^2$ is narrower than $y = x^2$ . This is called a dilation of factor $2$ from the $x$ -axis.

When $0 < a < 1$ : The parabola becomes broader (wider) than $y = x^2$ . For example, $y = \frac{1}{2}x^2$ is broader than $y = x^2$ .

When $a$ is negative: The parabola is reflected in the $x$ -axis, meaning it opens downward instead of upward. For example, $y = -2x^2$ is an upside-down parabola. This is called a reflection in the x-axis.

All these graphs still have their vertex at the origin $(0, 0)$ and are symmetric about the $y$ -axis.

Vertical translations (changing $k$ )

Adding or subtracting a constant $k$ to the function moves the parabola up or down.

For the function $y = x^2 + k$ :

When $k > 0$ : The graph shifts k units upward. For example, $y = x^2 + 1$ has its vertex at $(0, 1)$ .
When $k < 0$ : The graph shifts |k| units downward. For example, $y = x^2 - 2$ has its vertex at $(0, -2)$ .

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The axis of symmetry remains the $y$ -axis (the line $x = 0$ ), but the range changes. For $y = x^2 + k$ , the minimum value is $k$ instead of $0$ .

Horizontal translations (changing $h$ )

Changing $h$ in the function $y = (x - h)^2$ moves the parabola left or right.

For the function $y = (x - h)^2$ :

When $h > 0$ : The graph shifts h units to the right. For example, $y = (x - 2)^2$ has its vertex at $(2, 0)$ and axis of symmetry at $x = 2$ .
When $h < 0$ : The graph shifts |h| units to the left. For example, $y = (x + 3)^2$ (which is $y = (x - (-3))^2$ ) has its vertex at $(-3, 0)$ and axis of symmetry at $x = -3$ .

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Be careful with the signs! In the form $(x - h)^2$ , if we have $(x + 3)^2$ , this means $h = -3$ , so the graph moves 3 units to the left.

Combining transformations

By combining all these transformations, we can graph any quadratic in the form $y = a(x - h)^2 + k$ .

Key properties of $y = a(x - h)^2 + k$

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Key properties:

Vertex: The turning point is at $(h, k)$ .
Axis of symmetry: The line $x = h$ .
Width and direction: Determined by the value of $a$ (same as before).
Congruence: The graph has exactly the same shape as $y = ax^2$ , just moved to a different position.

To graph $y = a(x - h)^2 + k$ :

Start with the graph of $y = ax^2$
Translate it $h$ units horizontally (right if $h > 0$ , left if $h < 0$ )
Translate it $k$ units vertically (up if $k > 0$ , down if $k < 0$ )

Finding intercepts

To add more detail to your graph, you can find where it crosses the axes.

$y$ -intercept: Set $x = 0$ and solve for $y$ .

$x$ -intercepts: Set $y = 0$ and solve for $x$ . Note that some parabolas don't cross the $x$ -axis at all (when the vertex is above the $x$ -axis for upward-opening parabolas, or below for downward-opening ones).

Worked examples

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Worked Example: Sketching a Vertical Translation

Sketch the graph of $y = x^2 - 3$ .

Solution:

The graph of $y = x^2 - 3$ is obtained by taking the basic parabola $y = x^2$ and moving it 3 units down.

The vertex is at $(0, -3)$ .

The axis of symmetry is the line $x = 0$ (the $y$ -axis).

To find the $x$ -intercepts, set $y = 0$ :

0 = x^2 - 3

x^2 = 3

x = \pm\sqrt{3}

The $x$ -intercepts are at $x = \sqrt{3}$ and $x = -\sqrt{3}$ .

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Worked Example: Reflection and Translation

Sketch the graph of $y = -(x + 1)^2$ .

Solution:

This graph is obtained by starting with $y = x^2$ , reflecting it in the $x$ -axis (because of the negative sign), then moving it 1 unit to the left.

The vertex is at $(-1, 0)$ .

The axis of symmetry is the line $x = -1$ .

The $x$ -intercept is at $x = -1$ (the vertex is on the $x$ -axis).

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Worked Example: Combined Transformations with No x-intercepts

Sketch the graph of $y = 2(x - 1)^2 + 3$ .

Solution:

The graph is obtained from $y = 2x^2$ by translating 1 unit right and 3 units up.

The vertex is at $(1, 3)$ .

The axis of symmetry is the line $x = 1$ .

The graph is narrower than $y = x^2$ because $a = 2 > 1$ .

The range is $y \geq 3$ .

Finding the $y$ -intercept:

When $x = 0$ :

y = 2(0 - 1)^2 + 3

y = 2(1) + 3

y = 5

Finding the $x$ -intercepts:

The minimum value of $y$ is 3, which occurs at the vertex. Since the parabola opens upward and the vertex is above the $x$ -axis, the graph never crosses the x-axis. Therefore, there are no $x$ -intercepts.

We can verify this by setting $y = 0$ :

0 = 2(x - 1)^2 + 3

-3 = 2(x - 1)^2

-\frac{3}{2} = (x - 1)^2

Since we cannot take the square root of a negative number (in real numbers), there are no real solutions.

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Worked Example: Complete Analysis of a Parabola

Sketch the graph of $y = -(x + 1)^2 + 4$ .

Solution:

The vertex is at $(-1, 4)$ .

The axis of symmetry is the line $x = -1$ .

Finding the $y$ -intercept:

When $x = 0$ :

y = -(0 + 1)^2 + 4

y = -1 + 4

y = 3

Finding the $x$ -intercepts:

When $y = 0$ :

0 = -(x + 1)^2 + 4

(x + 1)^2 = 4

x + 1 = \pm 2

x = -1 + 2 = 1 \text{ or } x = -1 - 2 = -3

The $x$ -intercepts are at $x = 1$ and $x = -3$ .

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Key Points to Remember:

The graph of $y = x^2$ is a parabola with vertex at $(0, 0)$ and axis of symmetry along the $y$ -axis.
In the turning point form $y = a(x - h)^2 + k$ , the vertex is at (h, k) and the axis of symmetry is the line x = h.
The coefficient $a$ controls the width and direction: $|a| > 1$ makes the parabola narrower, $0 < |a| < 1$ makes it broader, and negative $a$ flips it upside down.
To find the $y$ -intercept, substitute $x = 0$ into the equation. To find $x$ -intercepts, set $y = 0$ and solve for $x$ (note that some parabolas have no $x$ -intercepts).
All parabolas of the form $y = a(x - h)^2 + k$ have the same shape as $y = ax^2$ , just translated to different positions.

Graphing Quadratics (VCE SSCE Mathematical Methods): Revision Notes