Linear, Quadratic and Cubic Graphs

Direction and Special Lines

• A positive gradient ($m > 0$) goes "uphill" from bottom-left to top-right, while a negative gradient ($m < 0$) goes "downhill" from top-left to bottom-right.
• A larger absolute value of $m$ creates a steeper line (e.g., $-5$ is steeper than $-4$).
• Horizontal lines are written as $y = k$ (where gradient is zero), and vertical lines are written as $x = k$ (where gradient is undefined).
• Parallel lines have identical gradients ($m_1 = m_2$), while perpendicular lines (Higher Tier) have gradients that are negative reciprocals ($m_2 = -1/m_1$).

Worked Example: Calculating Gradient

Calculate the gradient of the line passing through $A(7, -1)$ and $B(4, 5)$.

Step 1: State the formula for the gradient using two points.

$$m = \frac{y_2 - y_1}{x_2 - x_1}$$

Step 2: Substitute the coordinates, using brackets for negative numbers to avoid sign errors. $m = \frac{5 - (-1)}{4 - 7}$

Step 3: Calculate the final value. $m = \frac{6}{-3} = -2$

Sketching Linear Graphs

• To plot a graph, you must accurately calculate points using a table of values (using at least 3 points to verify the straight line).
• To sketch a graph, you only need to draw an approximate diagram showing the correct direction and labelling key intersections with the axes.

Worked Example: Sketching a Linear Graph

Sketch the graph of $y = -2x + 4$.

Step 1: Identify and label the $y$-intercept. The equation is in the form $y = mx + c$, so $c = 4$. Label the point $(0, 4)$ on the $y$-axis.

Step 2: Find the $x$-intercept by setting $y = 0$. $0 = -2x + 4$ $2x = 4$ $x = 2$, so label the point $(2, 0)$ on the $x$-axis.

Step 3: Draw a straight line through these points. Since $m = -2$, the line correctly slopes downhill from top-left to bottom-right.

Recognising and Interpreting Quadratic Graphs

You can easily predict a constant speed with a straight line, but understanding accelerating objects requires a curve.

• A quadratic function ($y = ax^2 + bx + c$) produces a smooth, symmetrical curve called a parabola.
• The shape is determined by the quadratic coefficient ($a$).
• If $a > 0$, the graph is u-shaped with a minimum turning point at its lowest value.
• If $a < 0$, the graph is n-shaped with a maximum turning point at its peak.

Key Features of a Parabola

• Roots: These are the $x$-intercepts where the graph crosses the $x$-axis (found by setting $y = 0$). A parabola can have 2, 1, or 0 roots.
• y-intercept: This occurs at $(0, c)$ where the graph crosses the $y$-axis.
• Line of symmetry: A vertical line exactly halfway between the roots, written as an equation $x = k$ (where $k$ is the $x$-coordinate of the turning point).
• For Higher Tier students, the turning point can be found by completing the square: $y = a(x + p)^2 + q$ gives a turning point at $(-p, q)$.

Sketching Quadratic Graphs

When asked to sketch a quadratic graph, follow these sequential steps:

1. Determine if it is u-shaped or n-shaped based on the sign of $x^2$.
2. Identify and label the $y$-intercept $(0, c)$.
3. Factorise or use the quadratic formula to find and label the roots.
4. Find the turning point (either midway between the roots or via completing the square) and label it.
5. Draw a single continuous smooth curve through your points.

Worked Example: Sketching a Quadratic Graph

Sketch the graph of $y = x^2 - 4x - 5$.

Step 1: Determine the shape and label the $y$-intercept. The coefficient of $x^2$ is positive ($a=1$), so the graph is a u-shaped parabola. The $y$-intercept is the constant term $c = -5$. Label the point $(0, -5)$ on the $y$-axis.

Step 2: Find and label the roots by setting $y = 0$. $0 = x^2 - 4x - 5$ Factorise the quadratic: $0 = (x - 5)(x + 1)$ The roots are $x = 5$ and $x = -1$. Label $(5, 0)$ and $(-1, 0)$ on the $x$-axis.

Step 3: Find and label the turning point. The line of symmetry is exactly halfway between the roots: $x = \frac{5 + (-1)}{2} = 2$. Substitute $x = 2$ back into the equation to find the $y$-coordinate: $y = (2)^2 - 4(2) - 5 = 4 - 8 - 5 = -9$. Label the minimum turning point at $(2, -9)$.

Step 4: Draw the curve. Draw a single continuous smooth u-shaped curve passing through $(-1, 0)$, $(0, -5)$, $(2, -9)$, and $(5, 0)$.

Recognising and Interpreting Cubic Graphs

Understanding how equations behave explains why some mathematical models rise, flatten out completely, and then rise again.

• A cubic function ($y = ax^3 + bx^2 + cx + d$) creates a distinctive "S-shape" or wiggle.
• It can have up to two turning points (a local maximum and a local minimum) or a single point of inflection (like in $y = x^3$) where the curvature flattens and changes direction.
• The graph will cross the $y$-axis at $(0, d)$ and can have 1, 2, or 3 roots.
• If a cubic function has a repeated root (e.g., a squared factor like $(x-1)^2$), the graph touches the $x$-axis at that point instead of crossing it.

End Behavior of Cubic Graphs

• End behavior describes the direction the graph travels as $x$ approaches positive or negative infinity.
• If the $x^3$ term is positive ($a > 0$), the graph starts bottom-left and ends top-right.
• If the $x^3$ term is negative ($a < 0$), the graph starts top-left and ends bottom-right.
• Higher Tier students must be able to sketch cubic graphs by factorising the equation to find the roots, while Foundation students focus on recognising the shapes and plotting from tables.

Worked Example: Sketching a Cubic Graph

Sketch the graph of $y = x^3 - 4x$.

Step 1: Identify and label the $y$-intercept. By setting $x = 0$, $y = (0)^3 - 4(0) = 0$. Label the point $(0, 0)$.

Step 2: Find and label the roots by setting $y = 0$. $0 = x^3 - 4x$ Factorise by taking out $x$: $0 = x(x^2 - 4)$ Factorise the difference of two squares: $0 = x(x - 2)(x + 2)$ The roots are $x = 0$, $x = 2$, and $x = -2$. Label $(0, 0)$, $(2, 0)$, and $(-2, 0)$ on the $x$-axis.

Step 3: Determine the end behavior and sketch the curve. The coefficient of $x^3$ is positive ($a = 1$), so the graph starts bottom-left and ends top-right. Draw a smooth curve starting from the bottom-left, passing up through $(-2, 0)$ to a local maximum, coming down through $(0, 0)$ to a local minimum, and going back up through $(2, 0)$ towards the top-right.