The curve above has a minimum point, so the shape resembles a 'happy face'.

The points $ (a, 0) $ and $ (b,0) $ are the x-intercepts. At these points, $ y = 0 $.

The point $ (0,c) $ is the y-intercept. At this point, $ x = 0 $.

The dotted line is the line of symmetry. It passes through the minimum point. If the x-intercepts are $ (a, 0) $ and $ (b,0) $, then

$$ \text{Line of symmetry, } x = \frac{a + b}{2} $$

The curve above has a maximum point, so the shape resembles a 'sad face'.

The points $ (a, 0) $ and $ (b,0) $ are the x-intercepts. At these points, $ y = 0 $.

The point $ (0,c) $ is the y-intercept. At this point, $ x = 0 $.

The dotted line is the line of symmetry. It passes through the maximum point. If the x-intercepts are $ (a, 0) $ and $ (b,0) $, then

$$ \text{Line of symmetry, } x = \frac{a + b}{2} $$

Example

The diagram below shows the graph of $ y = - x^2 + x + 2 $.

The graph cuts the $x$-axis at $A$ and $B$, cuts the $y$-axis at $C$. The maximum point is $D$.

Find the coordinates of $A$, $B$, $C$ and $D$.

Answer: $ A(-1, 0), B(2, 0), C(0, 2), D(0.5, 2.25) $

Solutions

1. Shape

• If the coefficient of $x^2$ is positive (i.e. $y = 2x^2 + 3x - 4$), then the graph is a minimum curve ($\cup$)
• If the coefficient of $x^2$ is negative (i.e. $y = -5x^2 + 6x + 7$), then the graph is a maximum curve ($\cap$)

2. Intercepts

• Solve for the $x$-intercept(s) by letting $ y = 0 $
• Solve for the $y$-intercept by letting $ x = 0 $

3. Line of symmetry

• The line of symmetry can be found by $x = $ $ \frac{a + b}{2} $, where $a$ and $b$ are the $x$-intercepts

4. Turning point

• The $x$-coordinate of the turning point is equal to the line of symmetry
• The $y$-coordinate of the turning point can be found by substituting the $x$-coordinate of the turning point into the equation of the curve

Example

Sketch the graph of $y = (3 - x)(x + 1)$. Indicate clearly the $x$-intercepts, $y$-intercept and the coordinates of the turning point.

Solutions

\begin{align*} y & = (3 - x)(x + 1) \\ y & = 3x + 3 - x^2 - x \\ y & = - x^2 + 2x + 3 \\ \\ \implies & \text{Maximum curve } \cap \\ \\ \text{Let } & y = 0, \\ 0 & = (3 - x)(x + 1) \end{align*} \begin{align*} 3 - x & = 0 && \text{ or } & x + 1 & = 0 \\ -x & = -3 &&& x & = -1 \\ x & = 3 \end{align*}

\begin{align*} x \text{-intercepts} & \text{ are } (-1,0), (3,0) \\ \\ \text{Let } & x = 0, \\ y & = (3 - 0)(0 + 1) \\ y & = 3 \\ \\ y \text{-inter} & \text{cept is } (0, 3) \\ \\ \text{Line of symmetry, } x & = \frac{-1 + 3}{2} \\ x & = 1 \\ \\ \text{Substitute } & x = 1 \text{ into eqn of curve,} \\ y & = (3 - 1)(1 + 1) \\ y & = 4 \\ \\ \text{Turning } & \text{point is } (1, 4) \end{align*}

$$ y = \pm (x - h)^2 + k $$

1. Shape

• The graph of $y = (x - h)^2 + k$ is a minimum curve ($\cup$)
• The graph of $y = -(x - h)^2 + k$ is a maximum curve ($\cap$)

2. Turning point

• The coordinates of the turning point is $ (h, k) $

3. $y$-intercept

• Solve for the $y$-intercept by letting $ x = 0 $

Note: In this form, we usually use the turning point, shape and y-intercept to sketch the graph. It is not always necessary to solve for the $x$-intercept(s). In fact, some quadratic curves do not meet the $x$-axis (see the next example).

Example

Sketch the graph of $y = (x - 1)^2 + 1$. Indicate clearly the $y$-intercept and the coordinates of the turning point.

Solutions

\begin{align*} y & = (x - 1)^2 + 1 \\ \\ \text{Mini} & \text{mum curve } \cup \\ \\ \text{Turn} & \text{ing point is } (1, 1) \\ \\ \text{Let } & x = 0, \\ y & = (0 - 1)^2 + 1 \\ y & = 2 \\ \\ y \text{-inte} & \text{rcept is } (0, 2) \end{align*}