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SL 2.3 · Sketch and interpret graphs from mathematical information

Learn to sketch and interpret graphs from mathematical information through clear examples and targeted practice.

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Functions

Turning equations, features, and context into a clear graph

A graph is a visual summary of how two quantities are related. To sketch one from mathematical information, you do not need to plot many points and hope they reveal the shape. Instead, identify what the equation tells you: which input values are allowed, where the graph meets the axes, how it behaves near important values, and whether familiar transformations are present. A good sketch shows these features clearly and labels important coordinates. In a contextual problem, it must also respect the situation and its units.

What you will learn

1. Prior knowledge: reading equations and coordinates

A point on a graph is written as (x,y)(x,y). The horizontal coordinate xx is the input and the vertical coordinate yy is the corresponding output. An equation such as y=f(x)y=f(x) gives the rule connecting them. The domain is the set of input values for which the rule is defined; a context may restrict it further.
To find an xx-intercept, set y=0y=0 and solve. To find a yy-intercept, set x=0x=0 and calculate the output, if 00 is in the domain. A root or zero is an input where the output is zero. These are the same locations as the graph's xx-intercepts.
For a quick symmetry check, compare the outputs at xx and −x-x. If f(−x)=f(x)f(-x)=f(x), the graph is symmetric about the yy-axis. This can reduce the work of sketching, but only if both inputs are in the domain.
x-intercept: f(x)=0x\text{-intercept: } f(x)=0

2. Build a sketch from mathematical features

Start by identifying the function family or a familiar parent graph. For example, y=x2y=x^2 is a parabola and y=1/xy=1/x is a reciprocal graph. Transformations shift, stretch, or reflect these familiar shapes. In y=(x−h)2+ky=(x-h)^2+k, the vertex is (h,k)(h,k); the graph of y=x2y=x^2 has been shifted horizontally by hh and vertically by kk.
Next find useful exact information. Intercepts often come from substitution or factorisation. For a quadratic, the axis of symmetry lies halfway between its two roots when both roots are known; it also passes through the vertex. For a reciprocal expression such as y=a/(x−h)+ky=a/(x-h)+k, the graph is undefined at x=hx=h, and its branches approach the lines x=hx=h and y=ky=k. These lines are asymptotes: the graph gets close to them but does not meet them in this model.
Plot a small number of points to confirm the shape and scale. Sketch smoothly rather than joining points with straight segments unless the relationship is defined piece by piece. Mark excluded inputs, asymptotes, intercepts, endpoints, and turning points where relevant. Do not add a point at an excluded input.
A graphing calculator is useful for checking the overall shape, locating intersections approximately, or noticing that a viewing window hides a feature. Choose a window that includes the domain and important coordinates. The calculator display supports the sketch; algebra and the stated domain explain why the features occur.
y=(x−h)2+ky=(x-h)^2+k

3. Interpret the graph, not just its shape

Interpretation means translating a visible feature into a statement about the quantities. An xx-intercept can represent when a measured amount becomes zero. A maximum or minimum can describe the greatest or least value shown, but its meaning depends on the domain and context. An increasing section means larger inputs correspond to larger outputs; a decreasing section means the opposite.
Always distinguish a mathematical graph from a contextual model. A formula may be defined for many real inputs, while the situation allows only some of them. Time might be restricted to non-negative values, and a count may use whole-number inputs. Include units when reporting contextual coordinates, and do not claim that the curve describes values outside the stated assumptions.
When information is given verbally or in a table, turn it into constraints before sketching. A starting value gives a point, a zero gives an intercept, and a stated maximum or minimum gives a turning point. Check that the proposed shape can satisfy all the information at once.

4. A practical exam-style routine

Read the question for the requested output: a sketch, an interpretation, or both. Record the domain and any units first. Then calculate features that can be found exactly and use those features to determine the curve's placement and shape. If the equation is unfamiliar, substitute a few sensible inputs and check the pattern.
For a technology check, enter the equation and select a window that shows the domain, intercepts, and any asymptotes or turning points. Compare the displayed graph with your analytical features. If they disagree, check the algebra, entered brackets, scale, and domain restrictions. For an exam sketch, make the essential features visible even when no exact scale is requested.

Worked example

A transformed quadratic

Sketch y=(x−2)2−4y=(x-2)^2-4 and state its vertex, intercepts, and symmetry.
  1. Read the transformation
    The expression is a shifted version of y=x2y=x^2. It is moved 2 units right and 4 units down, so its vertex is at (2,−4)(2,-4) and it opens upward.
    (2,−4)(2,-4)
  2. Find the horizontal intercepts
    Set the output equal to zero. The resulting equation factors into two linear factors, giving the two inputs where the curve crosses the horizontal axis.
    (x−2)2−4=0  ⇒  (x−4)x=0  ⇒  x=0,4(x-2)^2-4=0\;\Rightarrow\;(x-4)x=0\;\Rightarrow\;x=0,4
  3. Find the vertical intercept and symmetry
    Substitute x=0x=0 to obtain the vertical intercept. The vertex lies halfway between the two roots, so the symmetry axis is the vertical line through x=2x=2.
    y(0)=0,x=2y(0)=0,\qquad x=2
Answer: The graph is an upward-opening parabola with vertex (2,−4)(2,-4), intercepts (0,0)(0,0) and (4,0)(4,0), and axis of symmetry x=2x=2. Plot these points and draw a smooth symmetric curve.
Check: The vertex lies below both intercepts, as expected for an upward-opening parabola with a minimum.

Worked example

A reciprocal graph with excluded input

Sketch y=3x−1+2y=\frac{3}{x-1}+2. Identify its domain, asymptotes, and intercepts.
  1. Identify excluded input and asymptotes
    The denominator is zero when x=1x=1, so that input is excluded. In this translated reciprocal graph, the branches approach the vertical line through that excluded input and the horizontal line two units above the original reciprocal graph's centre level.
    x≠1,x=1,y=2x\ne 1,\qquad x=1,\qquad y=2
  2. Find the horizontal intercept
    Set y=0y=0 and solve. This gives the point where the graph crosses the horizontal axis.
    0=3x−1+2  ⇒  x=−120=\frac{3}{x-1}+2\;\Rightarrow\;x=-\frac{1}{2}
  3. Find the vertical intercept
    Substitute x=0x=0. This input is allowed, so the resulting point is on the graph.
    y(0)=3−1+2=−1y(0)=\frac{3}{-1}+2=-1
Answer: The domain is all real xx except 11. The asymptotes are x=1x=1 and y=2y=2. The intercepts are (−12,0)(-\frac{1}{2},0) and (0,−1)(0,-1). Sketch two reciprocal-shaped branches approaching both asymptotes without touching the excluded vertical line.
Check: For x>1x>1, the fraction is positive and the output is above 22. For x<1x<1, the output is below 22, consistent with the two branches.

Worked example

A contextual quadratic model

A ball's height above the ground, in metres, is modelled by h(t)=−t2+4t+5h(t)=-t^2+4t+5 for 0≤t≤50\leq t\leq 5, where tt is time in seconds. Sketch the graph and interpret its key features.
  1. Record the physical domain and starting height
    The stated interval limits the model to the flight from launch until the ball reaches the ground. At time zero, substitution gives the starting height.
    0≤t≤5,h(0)=50\leq t\leq 5,\qquad h(0)=5
  2. Find the maximum
    Complete the square to reveal the vertex. Since the squared term has a negative coefficient, the parabola opens downward and the vertex gives its maximum height.
    h(t)=−(t−2)2+9h(t)=-(t-2)^2+9
  3. Check when the ball reaches the ground
    Set the height equal to zero. The positive solution is in the stated time interval and represents the landing time; the negative solution is outside the model's domain. -t^2+4t+5=0 \Rightarrow (t-5)(t+1)=0 \Rightarrow t=5 or t=-1
Answer: Sketch the downward-opening parabola only for 0≤t≤50\leq t\leq 5. It starts at (0,5)(0,5), reaches a maximum height of 99 metres at t=2t=2 seconds, and returns to ground level at (5,0)(5,0). The negative root is not meaningful for this flight.
Check: The vertex time is within the flight interval, and the height is non-negative throughout the stated interval.

Common mistakes and how to avoid them

Plotting an intercept or point that is outside the stated domain.
Correction: Check every calculated input against the domain before including its point on the sketch.
Drawing a reciprocal graph through its vertical asymptote.
Correction: The function is undefined at that input; show branches approaching the asymptote without including a point on it.
Treating a calculator window as the full graph.
Correction: Adjust the viewing window and use the equation and domain to check what the graph should show.
Reporting a maximum without saying what its coordinates mean.
Correction: Give both coordinates and interpret them with the variable names and units.

Lesson summary

Check your understanding

Question 1

For y=(x+1)2−3y=(x+1)^2-3, what is the vertex?
  1. (1,−3)(1,-3)
  2. (−1,−3)(-1,-3)
  3. (−1,3)(-1,3)
  4. (0,−3)(0,-3)
Show answer and explanation
(−1,−3)(-1,-3)
The form (x−h)2+k(x-h)^2+k has vertex (h,k)(h,k). Here x+1=x−(−1)x+1=x-(-1), so the vertex is (−1,−3)(-1,-3).

Question 2

For y=2x+3−1y=\frac{2}{x+3}-1, which input is excluded?
  1. x=−3x=-3
  2. x=3x=3
  3. x=−1x=-1
  4. No input is excluded
Show answer and explanation
x=−3x=-3
The denominator is zero when x+3=0x+3=0, so x=−3x=-3 is excluded.

Question 3

A model is defined only for 0≤t≤40\leq t\leq 4. Its algebra gives a zero at t=−2t=-2. How should that zero be treated on the contextual sketch?
  1. Include it as the starting point.
  2. Ignore it as outside the stated domain.
  3. Replace it with t=2t=2.
  4. Use it as the maximum.
Show answer and explanation
Ignore it as outside the stated domain.
Although t=−2t=-2 solves the equation, it is outside the model's allowed interval and therefore is not a point on the contextual graph.

Key terms

Domain
The set of input values for which a function or model is considered.
Intercept
A point where a graph meets one of the coordinate axes.
Asymptote
A line that a graph approaches as its input or output changes.
Vertex
The turning point of a parabola; it is a maximum or minimum depending on the direction the parabola opens.
Transformation
A shift, stretch, or reflection that changes a familiar graph's position or shape.

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Published by DoAssignment. This AI-assisted lesson follows International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL, study topic SL 2.3. It is a study resource, not an official curriculum publication.

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