Graph Parabolas (HSC SSCE Mathematics Advanced): Revision Notes

Worked Example: Graphing a Monic Quadratic

Let's work through a complete example with $f(x) = x^2 - 4x + 3$.

Finding the x-intercepts

First, we set $f(x) = 0$ and use the discriminant to determine whether we can factorise or should use the quadratic formula.

Since the discriminant equals 4, which is a perfect square ($2^2 = 4$), we can factorise the equation.

The x-intercepts are $(1, 0)$ and $(3, 0)$.

Finding the y-intercept

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

$f(0) = 0^2 - 4 \times 0 + 3 = 3$

The y-intercept is at $(0, 3)$.

Finding the axis of symmetry and vertex

Using the formula for the axis of symmetry:

$x = -\frac{b}{2a} = -\frac{(-4)}{2 \times 1} = \frac{4}{2} = 2$

The axis of symmetry is the line $x = 2$.

To find the vertex, we substitute $x = 2$ into the function:

$f(2) = 2^2 - 4 \times 2 + 3$

$= 4 - 8 + 3$

$= -1$

The vertex is at $(2, -1)$.

Determining domain and range

The domain is $\mathbb{R}$ (all real numbers).

Since $a = 1 > 0$, the parabola is concave up with a minimum at $y = -1$. Therefore, the range is [$[-1, \infty)$].

Sketching the graph

We now plot all the key features: x-intercepts at $(1, 0)$ and $(3, 0)$, y-intercept at $(0, 3)$, vertex at $(2, -1)$, and the axis of symmetry at $x = 2$. The parabola opens upward.

Worked Example: Graphing a Non-Monic Quadratic

Let's work through $f(x) = 2x^2 + 4x - 3$.

Finding the x-intercepts

We set $f(x) = 0$ and calculate the discriminant to decide our approach.

Since the discriminant equals 40, which is not a perfect square, we need to use the quadratic formula.

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

$= \frac{-4 \pm \sqrt{4^2 - 4 \times 2 \times (-3)}}{2 \times 2}$

$= \frac{-4 \pm \sqrt{16 + 24}}{4}$

$= \frac{-4 \pm \sqrt{40}}{4}$

$= \frac{-4 \pm 2\sqrt{10}}{4}$

$= -1 \pm \frac{\sqrt{10}}{2}$

The x-intercepts are $\left(-1 + \frac{\sqrt{10}}{2}, 0\right)$ and $\left(-1 - \frac{\sqrt{10}}{2}, 0\right)$.

Finding the y-intercept

Substituting $x = 0$:

$f(0) = 2 \times 0^2 + 4 \times 0 - 3 = -3$

The y-intercept is at $(0, -3)$.

Finding the axis of symmetry and vertex

Using the axis of symmetry formula:

$x = -\frac{b}{2a} = -\frac{4}{2 \times 2} = -\frac{4}{4} = -1$

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

For the vertex, substitute $x = -1$:

$f(-1) = 2 \times (-1)^2 + 4 \times (-1) - 3$

$= 2 - 4 - 3$

$= -5$

The vertex is at $(-1, -5)$.

Determining domain and range

The domain is $\mathbb{R}$.

Since $a = 2 > 0$, the parabola is concave up with a minimum at $y = -5$. The range is [$[-5, \infty)$].

Sketching the graph

We plot the x-intercepts $\left(-1 + \frac{\sqrt{10}}{2}, 0\right)$ and $\left(-1 - \frac{\sqrt{10}}{2}, 0\right)$, the y-intercept $(0, -3)$, the vertex $(-1, -5)$, and draw the axis of symmetry at $x = -1$. The parabola opens upward and is steeper than a monic parabola because $a = 2 > 1$.