Graph Transformations, Rates of Change, and Circles (Higher)

Given a function $y=f(x)$, the following transformations produce new graphs:

Transformation	Equation	Effect on graph
Translate right by $a$	$y=f(x-a)$	Every point moves $a$ right
Translate left by $a$	$y=f(x+a)$	Every point moves $a$ left
Translate up by $b$	$y=f(x)+b$	Every point moves $b$ up
Translate down by $b$	$y=f(x)-b$	Every point moves $b$ down

Worked example — $f(x)=x^2$. Sketch $y=f(x-3)+2$.

Translate the graph of $y=x^2$ right by 3 and up by 2. The turning point moves from $(0,0)$ to $(3,2)$.

Key check: in $f(x-a)$, the sign in the bracket is opposite to the direction of movement. $f(x-3)$ moves right (not left).

Transformation	Equation	Effect
Reflect in $x$-axis	$y=-f(x)$	All $y$-values negate; graph flips vertically
Reflect in $y$-axis	$y=f(-x)$	All $x$-values negate; graph flips horizontally

Worked example — $f(x)=\sin x$. Sketch $y=-f(x)$ and $y=f(-x)$.

$y=-\sin x$: the sine wave flips upside down — peaks become troughs. The maximum of $y=-\sin x$ is $+1$ (at $x=270°$, since $-\sin (270°)=-(-1)=+1$); the minimum is $-1$ (at $x=90°$).

$y=\sin (-x)=-\sin x$: for $\sin$, reflecting in the $y$-axis gives the same result (sin is an odd function). For $\cos$: $\cos (-x)=\cos x$ — the graph is unchanged (cos is an even function).

Marking key features: state coordinates of any turning points, intercepts, or asymptotes after transformation.

Gradient at a point on a curve — estimated by drawing a tangent at that point and calculating the gradient of the tangent.

$\text{gradient}=\frac{\text{rise}}{\text{run}}\text{ of the tangent line}$

• On a distance-time graph: gradient = instantaneous speed
• On a velocity-time graph: gradient = acceleration
• On a financial graph: gradient = rate of change of value

Worked example — estimate the gradient of a curve at the point $(2,5)$ by drawing the tangent. The tangent passes through $(0,1)$ and $(4,9)$.

$\text{gradient}=\frac{9-1}{4-0}=2\text{ m/s²}$ (if this were a velocity-time graph, this is the acceleration at $t=2$)

Area under a graph:

• Under a velocity-time graph: area = distance travelled
• Under a distance-time graph: not physically meaningful as area
• Under a financial graph: may represent total cumulative value

Estimate irregular areas using the trapezium rule or by counting grid squares.

A circle with centre at the origin $(0,0)$ and radius $r$ has equation:

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

Worked examples:

$x^2+y^2=25$: centre $(0,0)$, radius $5$

$x^2+y^2=7$: centre $(0,0)$, radius $\sqrt{7}$

Does a point lie on the circle? Substitute the coordinates. If the equation is satisfied, the point is on the circle.

Point $(3,4)$ on $x^2+y^2=25$: $9+16=25$ ✓

Point $(2,3)$ on $x^2+y^2=25$: $4+9=13\ne 25$ — not on the circle.

Finding the radius: rearrange into the form $x^2+y^2=r^2$ and take the positive square root.

A radius from the origin to a point $(a,b)$ on the circle is perpendicular to the tangent at that point.

Method:

1. Find the gradient of the radius from $(0,0)$ to $(a,b)$: $m_r=\frac{b}{a}$
2. The tangent gradient is the negative reciprocal: $m_t=-\frac{a}{b}$
3. Use $y-b=m_t(x-a)$ to write the tangent equation.

Worked example — find the equation of the tangent to $x^2+y^2=25$ at the point $(3,4)$.

Gradient of radius: $m_r=\frac{4}{3}$

Tangent gradient: $m_t=-\frac{3}{4}$

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

Check: at $(3,4)$: $y=-\frac{9}{4}+\frac{25}{4}=\frac{16}{4}=4$ ✓

1. Transformations — sign direction error

$y=f(x-3)$ translates right by 3, not left. The rule is: $x-a$ moves the graph in the positive $x$-direction (right).

2. Gradient at a point — using the chord instead of the tangent

A chord connects two points on the curve; the tangent just touches it. Drawing a secant chord (not a tangent) and calculating its gradient gives an estimate for the average gradient, not the instantaneous gradient.

3. Circle equation — forgetting to square the radius

$x^2+y^2=5^2=25$, not $x^2+y^2=5$. The number on the right-hand side is $r^2$.

4. Tangent — using the same gradient as the radius

The tangent gradient is the negative reciprocal of the radius gradient, not the same. The radius from origin to $(3,4)$ has gradient $\frac{4}{3}$; the tangent gradient is $-\frac{3}{4}$.

Mistake	Correction
"$y=f(x+2)$ shifts right by 2"	$f(x+2)$ shifts left by 2 (replace $x$ with $x+2$: input decreases)
"Circle $x^2+y^2=9$ has radius 9"	Radius $=\sqrt{9}=3$
"Tangent at $(4,3)$ on $x^2+y^2=25$ has gradient $\frac{3}{4}$"	Radius gradient $=\frac{3}{4}$; tangent gradient $=-\frac{4}{3}$