Graph Transformations

Graph transformations are usually examined as a multiple-choice question or a short Section II part, often combined with another topic (a transformed hyperbola, exponential or circle). Year 12 extends the same ideas to trigonometric functions and modelling, so getting the rules right now pays off later.

• Reflections in the $x$-axis and $y$-axis.
• Translations horizontally and vertically.
• Dilations horizontally, vertically, and in both directions by the same factor.
• Applying combinations of these to the functions in the course (not trig functions, which are Year 12), finding the new equation, graph and key features, and describing the transformations.
• The order of transformations matters.
• Circles $(x-a{)}^2+(y-b{)}^2=r^2$, including completing the square from the general form.

The syllabus describes every transformation as a replacement in the equation. This works for any relation, including circles, and it explains why horizontal changes seem to go "the wrong way".

A negative $a$ or $b$ reverses the direction: replacing $x$ with $x+3$ is $x-(-3)$, a translation 3 units left.

A horizontal dilation stretches the graph away from the $y$-axis: every $x$-coordinate is multiplied by $k$ and the $y$-coordinates stay the same. A vertical dilation stretches away from the $x$-axis: every $y$-coordinate is multiplied by $l$. A factor between 0 and 1 compresses the graph towards the axis.

Replacing both $x$ with $\frac{x}{k}$ and $y$ with $\frac{y}{k}$ dilates the graph by factor $k$ in both directions: an enlargement if $k>1$, a reduction if $0<k<1$.

In function form

If you make $y$ the subject, replacing $y$ with $\frac{y-b}{l}$ turns $y=f(x)$ into $y=lf(x)+b$. Putting it all together:

• Vertical changes act outside the function and do what they look like: multiply $y$ by $l$, then add $b$.
• Horizontal changes act inside and do the opposite of what they look like: $f(2x)$ halves the $x$-coordinates (factor $\frac{1}{2}$), and $f(x-3)$ adds 3.

To transform a point, solve for where the old point lands. If $(p,q)$ is on $y=f(x)$, then on $y=lf\left(\frac{x-a}{k}\right)+b$ it moves to $(kp+a,lq+b)$.

Describe the transformations that map $y=\sqrt{x}$ onto $y=-2\sqrt{x+3}+1$, and state the domain and range of the new graph.

Compare with $y=lf(x-a)+b$: here $l=-2$, $a=-3$ and $b=1$.

1. Dilate vertically by factor 2.
2. Reflect in the $x$-axis.
3. Translate 3 units left.
4. Translate 1 unit up.

The domain changes from $x\ge 0$ to $x\ge -3$. The range changes from $y\ge 0$ to $y\le 0$ after the reflection, then to $y\le 1$ after the shift up. The endpoint moves from $(0,0)$ to $(-3,1)$.

Check with a point. $(4,2)$ is on $y=\sqrt{x}$. It moves to $(4-3,-2\times 2+1)=(1,-3)$. Substituting: $-2\sqrt{1+3}+1=-3$. Correct.

Order matters when a dilation or reflection and a translation act in the same direction.

• Vertically, $y=2f(x)+1$ means dilate by 2 then translate up 1. Translating first would give $y=2(f(x)+1)=2f(x)+2$, a different graph.
• Horizontally, the same thing happens inside the bracket.
• Horizontal and vertical changes don't affect each other, so they can be done in either order.

Worked example: a horizontal dilation and translation

Describe how to obtain $y=f(2x-6)$ from $y=f(x)$, and find the image of the point $(4,5)$.

Factorise inside the bracket: $f(2x-6)=f(2(x-3))$.

• Option 1: dilate horizontally by factor $\frac{1}{2}$, then translate 3 units right.
• Option 2: translate 6 units right (giving $f(x-6)$), then dilate horizontally by factor $\frac{1}{2}$ (replacing $x$ with $2x$ gives $f(2x-6)$).

Both are correct; the size of the translation depends on the order. For the image of $(4,5)$, the new graph has $y=5$ where $2x-6=4$:

The image is $(5,5)$. Check with option 1: $4\times \frac{1}{2}=2$, then $2+3=5$.

Track the key features through each transformation: intercepts, asymptotes, vertices, endpoints and centres.

• Parabola: $y=a(x-h{)}^2+k$ is $y=x^2$ dilated, possibly reflected, and translated so the vertex is $(h,k)$.
• Hyperbola: $y=\frac{c}{x-h}+k$ has asymptotes $x=h$ and $y=k$.
• Exponential: $y=a^{x-h}+k$ has asymptote $y=k$.
• Absolute value: $y=a|x-h|+k$ has its vertex at $(h,k)$.
• Cubic and semicircle: move the point of inflection or centre in the same way.

Example. $y=\frac{2}{x+1}-3$ is $y=\frac{2}{x}$ translated 1 left and 3 down. Its asymptotes are $x=-1$ and $y=-3$. The $y$-intercept is $\frac{2}{1}-3=-1$, and setting $y=0$ gives $x+1=\frac{2}{3}$, so the $x$-intercept is $-\frac{1}{3}$.

A circle with centre $(a,b)$ and radius $r$ is $x^2+y^2=r^2$ translated $a$ right and $b$ up:

Its domain is $[a-r,a+r]$ and its range is $[b-r,b+r]$.

If the equation is in general form $x^2+y^2+Dx+Ey+F=0$, complete the square in $x$ and in $y$ separately.

Worked example: centre and radius from the general form

Find the centre, radius, domain and range of $x^2+y^2+10x-4y+20=0$.

Centre $(-5,2)$, radius 3. Domain $[-8,-2]$, range $[-1,5]$.

Add the same numbers to both sides. If the right side comes out zero or negative, the equation is a single point or has no graph at all.

• Reading $f(x-3)$ as a shift left. It is 3 units right.
• Reading $f(2x)$ as a horizontal stretch by 2. It is a dilation by factor $\frac{1}{2}$.
• Translating $f(2x-6)$ by 6 units after dilating. Factorise to $f(2(x-3))$ first.
• Describing $y=-f(x)$ as a reflection in the $y$-axis. A minus sign outside reflects in the $x$-axis.
• Forgetting to add the completing-the-square numbers to the right side of a circle equation.
• Giving the centre of $(x+5{)}^2+(y-2{)}^2=9$ as $(5,-2)$, or the radius as 9.

• Describe questions want each transformation named with its direction and size, in an order that works, for example "dilate vertically by factor 2, then translate 1 unit up".
• Sketch questions: show the new intercepts, asymptotes, endpoints or centre with their coordinates.
• Check your new equation with one point you know on the original graph. It takes 20 seconds and catches most sign errors.
• In multiple choice, test each option with a known point instead of trying to reason about all four transformations at once.

• Working with functions (Year 11): the parent functions (lines, parabolas, cubics, hyperbolas, absolute value, circles) that get transformed here.
• Exponential and logarithmic functions (Year 11): $y=a^{-x}$ is a reflection of $y=a^x$, and log graphs are reflections in $y=x$.
• Further graph transformations and modelling (Year 12): the same rules applied to $y=k\sin (a(x+b))+c$ and to modelling real data.