Quadratic Functions Revision Notes for VCE SSCE Mathematical Methods

Quadratic Functions

Introduction

Quadratic functions are polynomial functions of degree $2$ . They can be expressed in different forms, and understanding how to work with these forms is essential for graphing and solving quadratic equations.

Transformations of parabolas

The basic parabola has the equation $y = x^2$ . We can transform this basic shape using dilations and translations to create any quadratic function.

Dilation from the x-axis

When we have a positive value of $a$ , the graph of $y = ax^2$ is obtained from the basic parabola $y = x^2$ by stretching or compressing it vertically.

If $a > 1$ , the parabola becomes narrower (vertical stretch)
If $0 < a < 1$ , the parabola becomes wider (vertical compression)
The larger the value of $a$ , the steeper the parabola

For example, comparing $y = 2x^2$ and $y = \frac{1}{2}x^2$ with $y = x^2$ shows how the dilation factor affects the width of the parabola.

Translation parallel to the x-axis

Horizontal translations shift the parabola left or right without changing its shape.

When $h > 0$ , the graph of $y = (x + h)^2$ is obtained by moving $y = x^2$ exactly $h$ units to the left (in the negative x-direction)
When $h < 0$ , the graph of $y = (x + h)^2$ moves $|h|$ units to the right (in the positive x-direction)

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This might seem counterintuitive at first. Remember: $y = (x + 2)^2$ shifts the graph $2$ units left, whilst $y = (x - 2)^2$ shifts it $2$ units right. The sign inside the parentheses is opposite to the direction of the shift.

Translation parallel to the y-axis

Vertical translations shift the parabola up or down.

When $k > 0$ , the graph of $y = x^2 + k$ moves $k$ units upward (in the positive y-direction)
When $k < 0$ , the graph of $y = x^2 + k$ moves $|k|$ units downward (in the negative y-direction)

Combinations of transformations

We can apply multiple transformations to create more complex quadratic functions. The general turning point form is $y = a(x - h)^2 + k$ , which represents:

A dilation by factor $a$ from the x-axis
A horizontal translation of $h$ units
A vertical translation of $k$ units

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Transformation Sequence

For $f(x) = 2(x - 2)^2 + 3$ , the transformations are applied as follows:

Start with $y = x^2$
Dilate by factor $2$ from the x-axis
Translate $2$ units in the positive x-direction (right)
Translate $3$ units in the positive y-direction (up)

Graphing quadratics in turning point form

The turning point form of a quadratic is $y = a(x - h)^2 + k$ . This form immediately reveals important features of the graph.

Key features:

The vertex (turning point) is located at $(h, k)$
The axis of symmetry is the vertical line $x = h$
If $a > 0$ , the parabola opens upward and has a minimum point at the vertex
If $a < 0$ , the parabola opens downward and has a maximum point at the vertex
The value of $|a|$ determines how narrow or wide the parabola is

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Worked Example: Sketching $y = 2(x - 1)^2 + 3$

From the equation, we can identify:

The vertex is at $(1, 3)$
The axis of symmetry is $x = 1$
Since $a = 2 > 0$ , the parabola opens upward and has a minimum point
The graph is narrower than $y = x^2$ because $a = 2 > 1$
The range is $(3, \infty)$

To find the y-axis intercept, substitute $x = 0$ :

y = 2(0 - 1)^2 + 3 = 2(1) + 3 = 5

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

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, this equation has no real solutions. Therefore, the graph has no x-axis intercepts. This makes sense because the minimum value is $3$ , which is above the x-axis.

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Worked Example: Sketching $y = -(x + 1)^2 + 4$

From the equation:

The vertex is at $(-1, 4)$
The axis of symmetry is $x = -1$
Since $a = -1 < 0$ , the parabola opens downward and has a maximum point

Finding the y-axis intercept (when $x = 0$ ):

y = -(0 + 1)^2 + 4 = -1 + 4 = 3

Finding the x-axis 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-axis intercepts are $1$ and $-3$ .

The axis of symmetry

For a quadratic function in polynomial form $y = ax^2 + bx + c$ , the axis of symmetry has the equation:

x = -\frac{b}{2a}

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This formula is extremely useful. The x-coordinate of the turning point is $-\frac{b}{2a}$ . To find the y-coordinate, substitute this x-value back into the quadratic equation.

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Worked Example: Finding the Turning Point

For the function $y = x^2 - 4x + 3$ , find the turning point and express in turning point form. State the minimum value and range.

Here $a = 1$ and $b = -4$ , so the axis of symmetry is:

x = -\frac{-4}{2(1)} = 2

When $x = 2$ :

y = (2)^2 - 4(2) + 3 = 4 - 8 + 3 = -1

The turning point is $(2, -1)$ .

In turning point form: $y = (x - 2)^2 - 1$

Since $a = 1 > 0$ , the parabola has a minimum point.

The minimum value is $-1$ and the range is $(-1, \infty)$ .

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Worked Example: Finding the Maximum

For the function $y = -2x^2 + 12x - 7$ , find the turning point and express in turning point form. State the maximum value and range.

Here $a = -2$ and $b = 12$ , so:

x = -\frac{12}{2(-2)} = 3

When $x = 3$ :

y = -2(3)^2 + 12(3) - 7 = -18 + 36 - 7 = 11

The turning point is $(3, 11)$ .

In turning point form: $y = -2(x - 3)^2 + 11$

Since $a = -2 < 0$ , the parabola has a maximum point.

The maximum value is $11$ and the range is $(-\infty, 11)$ .

Graphing quadratics in polynomial form

When a quadratic is in polynomial form $y = ax^2 + bx + c$ , we can sketch its graph by finding key features.

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Steps for graphing:

Step 1: Find the y-axis intercept (this is simply $c$ )

Step 2: Find the x-axis intercepts by solving $ax^2 + bx + c = 0$

Step 3: Find the axis of symmetry using $x = -\frac{b}{2a}$

Step 4: Find the turning point coordinates

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Worked Example: Graphing from Polynomial Form

Find the intercepts and turning point for $y = x^2 + x - 12$ , then sketch the graph.

Step 1: The y-axis intercept is $c = -12$

Step 2: To find x-axis intercepts, solve $x^2 + x - 12 = 0$ :

x^2 + x - 12 = 0

(x + 4)(x - 3) = 0

x = -4 \text{ or } x = 3

Step 3: The axis of symmetry is:

x = \frac{-4 + 3}{2} = -\frac{1}{2}

Step 4: When $x = -\frac{1}{2}$ :

y = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) - 12 = \frac{1}{4} - \frac{1}{2} - 12 = -12\frac{1}{4}

The turning point is $\left(-\frac{1}{2}, -12\frac{1}{4}\right)$ .

Completing the square

Completing the square is a technique for converting a quadratic from polynomial form to turning point form. This process reveals the vertex and makes graphing easier.

For $x^2 + bx + c$ :

Take half the coefficient of $x$ (which is $\frac{b}{2}$ ), then add and subtract its square $\left(\frac{b}{2}\right)^2 = \frac{b^2}{4}$ .

For $ax^2 + bx + c$ :

First factor out $a$ , then complete the square inside the brackets.

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Worked Example: Completing the Square

Write $f(x) = 2x^2 - 4x - 5$ in turning point form by completing the square, then sketch the graph.

f(x) = 2x^2 - 4x - 5

(add and subtract $\left(\frac{b}{2}\right)^2 = 1$ )

Now we can find the x-axis intercepts:

2x^2 - 4x - 5 = 0

2(x - 1)^2 - 7 = 0

(x - 1)^2 = \frac{7}{2}

x - 1 = \pm\sqrt{\frac{7}{2}}

x = 1 + \sqrt{\frac{7}{2}} \text{ or } x = 1 - \sqrt{\frac{7}{2}}

Key features:

y-axis intercept: $c = -5$
Turning point: $(1, -7)$
x-axis intercepts: $1 + \sqrt{\frac{7}{2}}$ and $1 - \sqrt{\frac{7}{2}}$

The quadratic formula

The solutions to the quadratic equation $ax^2 + bx + c = 0$ (where $a \neq 0$ ) are given by the quadratic formula:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

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This formula always works, even when the quadratic doesn't factorise easily. Notice that the axis of symmetry formula $x = -\frac{b}{2a}$ comes from the quadratic formula (it's the midpoint between the two solutions).

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Worked Example: Using the Quadratic Formula

Sketch the graph of $f(x) = -3x^2 - 12x - 7$ by finding the axis of symmetry, turning point, and using the quadratic formula to find x-axis intercepts.

The y-axis intercept is $c = -7$ .

Axis of symmetry:

x = -\frac{b}{2a} = -\frac{-12}{2(-3)} = -2

Turning point:

When $x = -2$ :

y = -3(-2)^2 - 12(-2) - 7 = -12 + 24 - 7 = 5

The turning point is $(-2, 5)$ .

x-axis intercepts:

-3x^2 - 12x - 7 = 0

Using the quadratic formula:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(-3)(-7)}}{2(-3)}

x = \frac{12 \pm \sqrt{144 - 84}}{-6}

x = \frac{12 \pm \sqrt{60}}{-6}

x = \frac{12 \pm 2\sqrt{15}}{-6}

x = -2 \pm \frac{1}{3}\sqrt{15}

So the x-intercepts are $-2 - \frac{1}{3}\sqrt{15}$ and $-2 + \frac{1}{3}\sqrt{15}$ .

The discriminant

The discriminant $\Delta$ of a quadratic polynomial $ax^2 + bx + c$ is:

\Delta = b^2 - 4ac

The discriminant tells us how many real solutions the equation $ax^2 + bx + c = 0$ has, which corresponds to the number of x-axis intercepts on the graph.

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Interpreting the discriminant:

If $\Delta > 0$ : two distinct real solutions (graph crosses x-axis twice)
If $\Delta = 0$ : one repeated real solution (graph touches x-axis once)
If $\Delta < 0$ : no real solutions (graph does not intersect x-axis)

For rational coefficients:

If $\Delta$ is a perfect square and $\Delta \neq 0$ : two rational solutions
If $\Delta = 0$ : one rational solution
If $\Delta$ is not a perfect square and $\Delta > 0$ : two irrational solutions

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Worked Example: Analyzing X-Axis Intersections

Without sketching, determine whether the graph of each function crosses, touches, or does not intersect the x-axis:

a) $f(x) = 2x^2 - 4x - 6$

Here $a = 2$ , $b = -4$ , $c = -6$

\Delta = b^2 - 4ac = (-4)^2 - 4(2)(-6) = 16 + 48 = 64 > 0

Since $\Delta > 0$ , the graph crosses the x-axis twice.

b) $f(x) = -4x^2 + 12x - 9$

Here $a = -4$ , $b = 12$ , $c = -9$

\Delta = (12)^2 - 4(-4)(-9) = 144 - 144 = 0

Since $\Delta = 0$ , the graph touches the x-axis once.

c) $f(x) = 3x^2 - 2x + 8$

Here $a = 3$ , $b = -2$ , $c = 8$

\Delta = (-2)^2 - 4(3)(8) = 4 - 96 = -92 < 0

Since $\Delta < 0$ , the graph does not intersect the x-axis.

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Worked Example: Finding Parameter Values

Find the values of $m$ for which $3x^2 - 2mx + 3 = 0$ has:

a) one solution

b) no solution

c) two distinct solutions

For the quadratic $3x^2 - 2mx + 3$ , the discriminant is:

\Delta = (-2m)^2 - 4(3)(3) = 4m^2 - 36

a) One solution: $\Delta = 0$

4m^2 - 36 = 0

m^2 = 9

m = \pm 3

b) No solution: $\Delta < 0$

4m^2 - 36 < 0

4m^2 < 36

m^2 < 9

-3 < m < 3

c) Two distinct solutions: $\Delta > 0$

4m^2 - 36 > 0

m^2 > 9

m > 3 \text{ or } m < -3

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Key Points to Remember:

Turning point form $y = a(x - h)^2 + k$ shows the vertex $(h, k)$ directly and whether the parabola opens up ( $a > 0$ ) or down ( $a < 0$ )
Axis of symmetry for $y = ax^2 + bx + c$ is always $x = -\frac{b}{2a}$
Completing the square converts polynomial form to turning point form by adding and subtracting $\left(\frac{b}{2}\right)^2$
Quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ finds x-intercepts for any quadratic
Discriminant $\Delta = b^2 - 4ac$ determines the number of solutions: $\Delta > 0$ (two solutions), $\Delta = 0$ (one solution), $\Delta < 0$ (no real solutions)

Quadratic Functions (VCE SSCE Mathematical Methods): Revision Notes