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Year 12 Maths - Methods (Unit 3 & Unit 4) Transformations

Determining the rule for a function from its graph

20 practice questions 0 video lessons Theory + worked examples
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Theory

Determining the rule of a function from its graph means recognising the base family from the shape and reading the transformations off key features. Every graph here is a transformed standard function \(y=a\,f\big(n(x-h)\big)+k\): read \(h\) and \(k\) from the vertex, endpoint or asymptotes, find \(a\) (and where needed the base) from one more marked point, then write and check the equation.

Each graph in this course is a transformation of a standard "parent" function such as \(y=x^{2}\), \(y=x^{3}\), \(y=\dfrac{1}{x}\), \(y=\sqrt{x}\) or \(y=b^{x}\). The general transformed rule is \(y=a\,f\big(n(x-h)\big)+k\), where \(h\) and \(k\) are horizontal and vertical shifts, \(a\) is a vertical dilation (and reflection if \(a<0\)) and \(n\) is a horizontal dilation. To recover the rule you decide which parent function fits the shape, then pin the constants to features you can read off the graph.

The shifts \(h\) and \(k\) are the coordinates of the family's invariant feature. For a parabola this is the vertex \((h,k)\); for a cubic it is the point of inflection; for a square root it is the endpoint; and for a hyperbola or an exponential the lines \(x=h\) and \(y=k\) are the vertical and horizontal asymptotes. Reading that feature gives \(h\) and \(k\) immediately.

With \(h\) and \(k\) fixed, the dilation factor \(a\) is found by substituting one further marked point into the partial rule and solving. If a horizontal dilation or an unknown base is also present, use a second point to get a second equation. Finally, always check the completed rule against another point or intercept on the graph.

Key idea. Read \(h,k\) from the invariant feature (vertex / endpoint / point of inflection / asymptotes), then solve for \(a\) using a known point: substitute into \(y=a\,f\big(n(x-h)\big)+k\) and make \(a\) the subject.
Parabola with vertex (2,-1)A parabola opening upward with its vertex marked at (2,-1) and passing through the point (0,3), used to determine the rule y=(x-2)^2-1. x y vertex (2, -1) (0, 3)
Vertex at \((2,-1)\) gives \(h=2,\ k=-1\); the point \((0,3)\) fixes \(a\), so \(y=(x-2)^{2}-1\)
Hyperbola with asymptotes x=2 and y=1A hyperbola with a vertical asymptote at x=2 and horizontal asymptote at y=1, passing through (3,4), used to find the rule y=3/(x-2)+1. x y (3, 4) y = 1 x = 2
Asymptotes \(x=2,\ y=1\) give \(h=2,\ k=1\); the point \((3,4)\) fixes \(a=3\), so \(y=\dfrac{3}{x-2}+1\)

The general transformed rule — \(a\) dilates vertically (and reflects if \(a<0\)), \(n\) dilates horizontally, and \((h,k)\) shifts:

\[y=a\,f\big(n(x-h)\big)+k\]
y=af(n(xh))+k

Vertex form of a parabola and point-of-inflection form of a cubic — \((h,k)\) is the marked feature:

\[y=a(x-h)^{2}+k \qquad y=a(x-h)^{3}+k\]
y=a(xh)2+k

Hyperbola and square root — here \(x=h\) is a vertical asymptote (hyperbola) and \((h,k)\) is the endpoint (square root):

\[y=\dfrac{a}{x-h}+k \qquad y=a\sqrt{x-h}+k\]
y=axh+k

Exponential — \(y=k\) is the horizontal asymptote; two points determine \(a\) and the base \(b\):

\[y=a\,b^{\,x}+k \qquad y=a\,e^{\,n(x-h)}+k\]
y=abx+k
Reading the feature. A vertex, endpoint or point of inflection gives \((h,k)\) directly; a hyperbola or exponential gives \(k\) from the horizontal asymptote (and \(h\) from the vertical asymptote). Then one more point solves for \(a\).

How to recover a rule from a graph

  1. Identify the base family. Match the shape to \(x^{2}\), \(x^{3}\), \(\dfrac{1}{x}\), \(\sqrt{x}\) or \(b^{x}\) — a single turning point, an S-shape, two asymptotic branches, an endpoint, or one horizontal asymptote.
  2. Read \(h\) and \(k\). Take them from the invariant feature: the vertex, point of inflection, endpoint, or the asymptotes \(x=h\) and \(y=k\).
  3. Write the partial rule. Substitute \(h\) and \(k\), leaving \(a\) (and, if needed, \(n\) or the base) as unknowns, e.g. \(y=a(x-h)^{2}+k\).
  4. Solve for \(a\). Substitute one more marked point and make \(a\) the subject. If two constants are unknown (an exponential base, say), use two points to form two equations.
  5. Check. Test the finished rule against another point, intercept or the orientation of the curve.
Orientation check. Decide the sign of \(a\) from the shape before you compute: a downward parabola, a falling square root, or a flipped exponential all need \(a<0\). If your algebra gives the wrong sign, re-read the feature.
Example 1 — parabola from its vertex
A parabola has vertex \((2,-1)\) and passes through \((0,3)\). Find its rule.
Solution
Vertex form — read \(h=2,\ k=-1\):
\(y\)\(=\)\(a(x-2)^{2}-1\)
Substitute the point \((0,3)\) and solve for \(a\):
\(3\)\(=\)\(a(0-2)^{2}-1\)
\(3\)\(=\)\(4a-1\)
\(a\)\(=\)\(1\)
\(\therefore\) \(y=(x-2)^{2}-1\)
Parabola vertex (2,-1) through (0,3)A parabola with vertex (2,-1) passing through (0,3). x y (2, -1) (0, 3)
y=(x2)21
Example 2 — square root from its endpoint
A square-root curve starts at the endpoint \((1,2)\) and passes through \((5,0)\). Find its rule.
Solution
Endpoint form — read \(h=1,\ k=2\):
\(y\)\(=\)\(a\sqrt{x-1}+2\)
Substitute \((5,0)\) — note \(\sqrt{5-1}=2\):
\(0\)\(=\)\(a\sqrt{4}+2\)
\(0\)\(=\)\(2a+2\)
\(a\)\(=\)\(-1\)
\(a<0\) matches a curve that falls from the endpoint:
\(\therefore\) \(y=-\sqrt{x-1}+2\)
y=x1+2
Example 3 — hyperbola from its asymptotes
A hyperbola has asymptotes \(x=2\) and \(y=1\) and passes through \((3,4)\). Find its rule.
Solution
Asymptotes give \(h=2,\ k=1\):
\(y\)\(=\)\(\dfrac{a}{x-2}+1\)
Substitute \((3,4)\) and solve for \(a\):
\(4\)\(=\)\(\dfrac{a}{3-2}+1\)
\(4\)\(=\)\(a+1\)
\(a\)\(=\)\(3\)
\(\therefore\) \(y=\dfrac{3}{x-2}+1\)
y=3x2+1
Example 4 — exponential from asymptote and two points
An exponential curve has horizontal asymptote \(y=1\) and passes through \((0,3)\) and \((1,5)\). Find its rule in the form \(y=a\,b^{x}+1\).
Solution
Asymptote gives \(k=1\); two unknowns \(a,b\) remain:
\(y\)\(=\)\(a\,b^{x}+1\)
Use \((0,3)\) — since \(b^{0}=1\):
\(3\)\(=\)\(a(1)+1\)
\(a\)\(=\)\(2\)
Now use \((1,5)\) with \(a=2\):
\(5\)\(=\)\(2b+1\)
\(b\)\(=\)\(2\)
Check \((1,5)\): \(2(2)^{1}+1=5\) ✓
\(\therefore\) \(y=2\cdot 2^{x}+1\)
y=2·2x+1

Common pitfalls

Watch the sign of \(h\). The rule contains \((x-h)\), so a feature at \(x=2\) gives \((x-2)\), not \((x+2)\). A left shift to \(x=-3\) gives \((x+3)\).
Do not forget a reflection. If the curve opens or points the "wrong" way — a downward parabola, a falling square root — then \(a\) is negative. Decide the sign from the shape before solving.
Find \(a\) from a point, not by eye. Never guess the dilation from how steep the curve looks; substitute a marked point and solve. An exponential or hyperbola needs the asymptote for \(k\) and a point for \(a\).

Frequently asked questions

How do you find the rule of a function from its graph?

Recognise the base family from the shape, read \(h\) and \(k\) from a key feature (vertex, endpoint or asymptotes), write \(y=a\,f\big(n(x-h)\big)+k\), then substitute one more point to find \(a\) and write the full equation.

How do you read h and k from a graph?

They are the coordinates of the invariant feature: the vertex of a parabola, the point of inflection of a cubic, the endpoint of a square root, or the asymptotes \(x=h\) and \(y=k\) of a hyperbola or exponential.

How do you find the value of a?

With \(h\) and \(k\) known, substitute one further marked point into the rule and solve. For \(y=a(x-2)^{2}-1\) through \((0,3)\): \(3=4a-1\), so \(a=1\).

How do you tell which base function a graph comes from?

A single turning point is a parabola; an S-shape through a point of inflection is a cubic; two branches with two asymptotes is a hyperbola; a curve starting at an endpoint is a square root; and one horizontal asymptote with growth or decay is an exponential.

What does a negative value of a mean?

\(a<0\) reflects the graph in a horizontal line through the feature, so the curve opens or points the opposite way. Always check the orientation against the sign you obtain for \(a\).

How do you find the rule of an exponential graph?

Read the asymptote for \(k\) in \(y=a\,b^{x}+k\), then substitute two points to solve for \(a\) and \(b\). Asymptote \(y=1\) through \((0,3)\) and \((1,5)\) gives \(y=2\cdot 2^{x}+1\).