Graphs

Every straight line can be written in the form:

$y = mx + c$

where $m$ is the gradient (steepness) and $c$ is the $y$-intercept (where the line crosses the $y$-axis).

Calculating gradient:

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

A positive gradient slopes upward left-to-right; a negative gradient slopes downward.

Finding the equation from two points:

Through $(1, 3)$ and $(3, 7)$:

$m = \frac{7 - 3}{3 - 1} = \frac{4}{2} = 2$

Substitute into $y = mx + c$: $3 = 2(1) + c \implies c = 1$, so $y = 2x + 1$.

Parallel and perpendicular lines:

• Parallel lines have equal gradients.
• Perpendicular lines have gradients that are negative reciprocals: $m_1 \times m_2 = -1$.

The graph of $y = ax^2 + bx + c$ is a parabola.

• If $a > 0$: U-shaped (minimum turning point)
• If $a < 0$: ∩-shaped (maximum turning point)

Key features:

• Roots (x-intercepts): where $y = 0$; found by factorising or the quadratic formula
• $y$-intercept: set $x = 0$; gives $(0, c)$
• Turning point (vertex): at $x = -\dfrac{b}{2a}$; substitute back for $y$
• Line of symmetry: $x = -\dfrac{b}{2a}$

Quadratic formula (when factorising is not straightforward):

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

Discriminant $= b^2 - 4ac$:

• $> 0$: two distinct real roots (crosses $x$-axis twice)
• $= 0$: one repeated root (touches $x$-axis)
• $< 0$: no real roots (does not cross $x$-axis)

You must be able to recognise and sketch each of these graph families.

Cubic $y = ax^3 + \ldots$:

• Positive $a$: rises left-to-right with an S-shape
• Has up to 3 roots and up to 2 turning points

Reciprocal $y = \dfrac{k}{x}$ ($x \neq 0$):

• Two branches in opposite quadrants (for $k > 0$: Q1 and Q3)
• Asymptotes along both axes — the curve never touches them

Exponential $y = k^x$ (for $k > 1$):

• Always positive; crosses the $y$-axis at $(0, 1)$
• Increases rapidly; never touches the $x$-axis (asymptote at $y = 0$)
• For $0 < k < 1$: decreasing exponential (decay)

Trigonometric graphs (arguments in degrees):

• $y = \sin x$: period 360°, amplitude 1, passes through $(0,0)$
• $y = \cos x$: period 360°, amplitude 1, passes through $(0,1)$
• $y = \tan x$: period 180°, asymptotes at $x = 90°, 270°, \ldots$

Transformations describe how a graph moves or flips when the function is modified.

Transformation	Effect on graph
$y = f(x) + a$	Translate by $\begin{pmatrix} 0 \\ a \end{pmatrix}$ (up by $a$)
$y = f(x) - a$	Translate by $\begin{pmatrix} 0 \\ -a \end{pmatrix}$ (down by $a$)
$y = f(x + a)$	Translate by $\begin{pmatrix} -a \\ 0 \end{pmatrix}$ (left by $a$)
$y = f(x - a)$	Translate by $\begin{pmatrix} a \\ 0 \end{pmatrix}$ (right by $a$)
$y = -f(x)$	Reflect in the $x$-axis
$y = f(-x)$	Reflect in the $y$-axis

Example: The graph of $y = \sin x$ is transformed to $y = \sin(x - 30°) + 2$.

This is a translation of $\begin{pmatrix} 30 \\ 2 \end{pmatrix}$ — 30° to the right and 2 units up.

Graphs are used to represent real-world relationships. You must be able to read off values, find rates of change, and interpret the meaning of gradient and area.

Distance-time graphs:

• Gradient = speed
• Horizontal line = stationary
• Steeper line = faster speed

Velocity-time graphs:

• Gradient = acceleration (negative gradient = deceleration)
• Area under graph = distance travelled

Estimating gradient of a curve (Higher):

Draw a tangent at the point of interest (a straight line touching the curve at that point without crossing it). Calculate the gradient of the tangent using two points on it.

Estimating area under a curve (Higher):

Use the trapezium rule: divide the area into trapeziums of equal width $h$:

$A \approx \frac{h}{2}\left[y_0 + 2(y_1 + y_2 + \ldots + y_{n-1}) + y_n\right]$

Or count squares on graph paper as an estimate.

Equation of a circle with centre $(0, 0)$ and radius $r$:

$x^2 + y^2 = r^2$

Example: $x^2 + y^2 = 25$ is a circle of radius 5 centred at the origin.

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Finding the equation of a tangent at a point (Higher):

A tangent to a circle is perpendicular to the radius at the point of contact.

Method: For the circle $x^2 + y^2 = r^2$ and the point $P(a, b)$ on the circle:

1. The radius $OP$ has gradient $\dfrac{b}{a}$.
2. The tangent is perpendicular, so its gradient is $-\dfrac{a}{b}$.
3. Use $y - b = m(x - a)$ with $m = -\dfrac{a}{b}$.

Example: Find the tangent to $x^2 + y^2 = 25$ at $(3, 4)$:

Gradient of radius $= \dfrac{4}{3}$; tangent gradient $= -\dfrac{3}{4}$

$y - 4 = -\frac{3}{4}(x - 3) \implies 4y - 16 = -3x + 9 \implies 3x + 4y = 25$