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Functions & Graphs — Grade 11

Learn the horizontal shift parameter p on parabolas, hyperbolas and exponential graphs, and take your first steps into average gradient. Notes, past papers and a quiz — everything for this topic is one click away.

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Grade 11 CAPS Mathematics

Functions & Graphs

The parameter \(p\), and your first look at average gradient.

What You Already Know From Grade 10

Grade 10 taught you two parameters. This lesson adds a third.

The parameter \(q\)

A vertical shift. In \(y=a\cdot f(x)+q\), the graph moves up if \(q>0\) and down if \(q<0\).

The parameter \(a\)

Controls reflection and stretch. A negative \(a\) flips the graph; a larger \(|a|\) stretches it, a smaller \(|a|\) compresses it.

The three families

You can already sketch \(y=a x^2+q\), \(y=\dfrac{a}{x}+q\), and \(y=ab^x+q\).

What's missing

None of these move the graph left or right yet. That's the job of the new parameter \(p\).

The New Parameter: \(p\) (Horizontal Shift)

CAPS writes this parameter as \(+p\), which makes the shift direction the opposite of what most learners expect.

Two parabolas showing the effect of the parameter p The graph of y equals x squared with turning point at the origin, and the graph of y equals open bracket x plus 3 close bracket squared, shifted 3 units to the left with turning point at negative 3, 0. x y=x² — TP (0,0) y=(x+3)² — TP (-3,0)
Reading the diagram: \(y=(x+3)^2\) is \(y=x^2\) shifted 3 units left — even though the parameter is a positive \(+3\).
General forms (CAPS notation)
  • Parabola: \(y=a(x+p)^2+q\)
  • Hyperbola: \(y=\dfrac{a}{x+p}+q\)
  • Exponential: \(y=ab^{x+p}+q\ (b>0,\,b\neq1)\)
Finding the shift

Set the bracket to zero: \(x+p=0 \Rightarrow x=-p\). That's where the key feature (turning point or asymptote) sits.

Common mistake
Because the form is \(x+p\) (not \(x-p\)), a positive \(p\) shifts the graph left, and a negative \(p\) shifts it right — the opposite of what \(q\)'s sign does vertically. Always solve \(x+p=0\) rather than guessing.
Worked example Level 1

Given \(f(x)=2(x+3)^2-4\), determine the turning point and the equation of the axis of symmetry.

Show solution
  1. 1Compare with \(y=a(x+p)^2+q\): here \(a=2\), \(p=3\), \(q=-4\).
  2. 2Find the turning point's x-coordinate: \(x+3=0 \Rightarrow x=-3\).
  3. 3So the turning point is \(\boxed{(-3,-4)}\), and the axis of symmetry is \(\boxed{x=-3}\).
Worked example Level 3

The graph below has a turning point at \((-1,4)\) and passes through \((0,2)\). Determine the equation in the form \(y=a(x+p)^2+q\).

A downward parabola with turning point negative one comma four and y-intercept zero comma two A parabola opening downward with its turning point at negative one comma four, passing through the point zero comma two on the y-axis. (-1,4) (0,2)
Reading the diagram: the turning point \((-1,4)\) fixes \(p\) and \(q\); the second point \((0,2)\) is then substituted in to solve for \(a\).
Show solution
  1. 1The turning point is \((-p,q)\). Reading it off the graph, \((-p,q)=(-1,4)\), so \(p=1\) and \(q=4\).
  2. 2Substitute into \(y=a(x+p)^2+q\): \(y=a(x+1)^2+4\).
  3. 3Use the second point \((0,2)\) to solve for \(a\): \(2=a(0+1)^2+4 \Rightarrow 2=a+4 \Rightarrow a=-2\).
  4. 4So the equation is \(\boxed{y=-2(x+1)^2+4}\).
Quick Check

For \(y=a(x+p)^2+q\), the turning point is:

What is the turning point of \(f(x)=(x+4)^2-1\)?

Horizontal Shift on the Hyperbola

The same rule applies: the vertical asymptote moves to \(x=-p\).

Comparing y equals one over x with y equals one over open bracket x plus two close bracket minus one The base hyperbola y equals one over x with asymptotes at the origin, shown dashed, and the shifted hyperbola y equals one over x plus two minus one with asymptotes x equals negative two and y equals negative one, shown solid, two units to the left and one unit down. y=1/x y=1/(x+2)-1
Reading the diagram: \(y=\dfrac{1}{x+2}-1\) is \(y=\dfrac{1}{x}\) shifted 2 units left and 1 unit down — both asymptotes move with it.
Form
\[y=\dfrac{a}{x+p}+q\]
Asymptotes

Vertical: \(x=-p\) (set the denominator to zero). Horizontal: \(y=q\) (unaffected by \(p\)).

Why the vertical asymptote moves

The break happens wherever the denominator is zero. \(x+p=0\) solves to \(x=-p\), not \(x=p\) — the same sign flip as the parabola.

Domain and range

Domain: \(x\in\mathbb{R},\,x\neq-p\). Range: \(y\in\mathbb{R},\,y\neq q\) — both restrictions simply shift along with the asymptotes.

Worked example Level 1

Given \(g(x)=\dfrac{3}{x+2}-1\), write down the equations of the asymptotes.

Show solution
  1. 1Vertical asymptote: set \(x+2=0 \Rightarrow x=-2\).
  2. 2Horizontal asymptote: \(y=q=-1\).
  3. 3So \(\boxed{x=-2}\) and \(\boxed{y=-1}\).
Worked example Level 3

Given \(k(x)=\dfrac{-5}{x-3}+2\), determine the equations of the asymptotes, and state the domain and range.

Show solution
  1. 1Match to the CAPS form \(x+p\): \(x-3\) means \(p=-3\) — a negative \(p\), so be careful with the sign.
  2. 2Vertical asymptote: set \(x-3=0 \Rightarrow x=3\).
  3. 3Horizontal asymptote: \(y=q=2\).
  4. 4Domain: \(\boxed{x\in\mathbb{R},\,x\neq3}\). Range: \(\boxed{y\in\mathbb{R},\,y\neq2}\).
Worked example Level 4

A hyperbola of the form \(y=\dfrac{a}{x+p}+q\) has asymptotes \(x=5\) and \(y=-2\), and passes through the point \((6,1)\). Determine the values of \(a\), \(p\) and \(q\).

Show solution
  1. 1The vertical asymptote gives \(p\): \(-p=5 \Rightarrow p=-5\).
  2. 2The horizontal asymptote gives \(q\) directly: \(q=-2\).
  3. 3Substitute the point \((6,1)\) into \(y=\dfrac{a}{x-5}-2\): \(1=\dfrac{a}{6-5}-2 \Rightarrow 1=a-2 \Rightarrow a=3\).
  4. 4So \(\boxed{a=3,\ p=-5,\ q=-2}\), giving \(y=\dfrac{3}{x-5}-2\).
Quick Check

For \(y=\dfrac{a}{x+p}+q\), finding the vertical asymptote requires:

For \(g(x)=\dfrac{6}{x+1}-4\), the horizontal asymptote is:

Horizontal Shift on the Exponential

The horizontal asymptote still only depends on \(q\) — \(p\) only moves the curve sideways.

Comparing y equals two to the power x with y equals two to the power open bracket x plus two close bracket minus three The base exponential y equals two to the power x with asymptote y equals zero, shown dashed, and the shifted exponential y equals two to the power x plus two minus three with asymptote y equals negative three, shown solid, two units to the left and three units down. y=0 y=-3 y=2ₓ y=2ₓ⁺²-3
Reading the diagram: \(y=2^{x+2}-3\) is \(y=2^x\) shifted 2 units left and 3 units down — the asymptote moves with the vertical shift only, never with \(p\).
Form
\[y=ab^{x+p}+q\]
What each parameter does
  • \(a\): reflection and stretch
  • \(p\): horizontal shift (left if \(p>0\))
  • \(q\): horizontal asymptote \(y=q\), and vertical shift
Why \(p\) never touches the asymptote

The asymptote comes from \(b^{x+p}\to0\) as \(x\to-\infty\) — that limit is \(0\) regardless of \(p\), so the curve always settles at \(y=q\), no matter how far it's shifted sideways.

Worked example Level 1

Given \(h(x)=2^{x+1}-3\), determine the equation of the asymptote and the y-intercept.

Show solution
  1. 1The asymptote depends only on \(q\): \(\boxed{y=-3}\) (the shift by \(p\) does not affect it).
  2. 2For the y-intercept, let \(x=0\): \(h(0)=2^{0+1}-3=2^1-3=2-3=-1\).
  3. 3So the y-intercept is \(\boxed{(0,-1)}\).
Quick Check

For \(y=ab^{x+p}+q\), the horizontal asymptote \(y=q\) is affected by:

For \(h(x)=2^{x+3}-5\), the y-intercept is:

All Three Parameters Together

Every Grade 11 sketching question is really this table, applied to one of the three families.

ParameterControlsRule
\(a\)Reflection & stretch\(a<0\) reflects; \(|a|>1\) stretches, \(0<|a|<1\) compresses
\(p\)Horizontal shiftSolve \(x+p=0\): shift is left if \(p>0\), right if \(p<0\)
\(q\)Vertical shift / asymptoteUp if \(q>0\), down if \(q<0\)
Worked example Level 3

Describe, in words, how the graph of \(y=-2(x+1)^2+5\) is obtained from \(y=x^2\).

Show solution
  1. 1\(a=-2\): reflect in the x-axis, and stretch vertically by a factor of 2.
  2. 2\(p=1\): since \(x+1=0\Rightarrow x=-1\), shift 1 unit left.
  3. 3\(q=5\): shift 5 units up.
  4. 4The new turning point is \(\boxed{(-1,5)}\), and the parabola opens downward.
Quick Check

Which parameter never moves an asymptote or turning point on its own — it only changes shape (stretch/reflection)?

For \(y=-2(x+5)^2-3\), which statement is correct?

Average Gradient: An Intuitive Introduction

A curve's steepness keeps changing — so instead of one gradient, we measure the gradient between two points.

A curve with a secant line joining two points A and B A curve with point A at one, negative one and point B at four, eight, joined by a straight secant line whose gradient is the average gradient between A and B. A(1,-1) B(4,8)
Key idea: the blue line (the secant) joins A and B directly. Its gradient is the average gradient of the curve between those two points — it ignores what the curve does in between.
Formula
\[m=\dfrac{y_2-y_1}{x_2-x_1}\]

The same gradient formula you already know — just applied to two points that happen to lie on a curve instead of a straight line.

Why it's only an estimate

The closer together the two points are, the closer this average gradient gets to the curve's actual steepness at that location. (In Grade 12, this idea becomes the derivative.)

Worked example Level 1

For \(f(x)=x^2-2x\), determine the average gradient between \(x=1\) and \(x=4\).

Show solution
  1. 1Find the two y-values: \(f(1)=1^2-2(1)=1-2=-1\).
  2. 2\(f(4)=4^2-2(4)=16-8=8\).
  3. 3Apply the formula: \(m=\dfrac{8-(-1)}{4-1}=\dfrac{9}{3}\).
  4. 4So the average gradient is \(\boxed{3}\).
Quick Check

The average gradient between two points on a curve is calculated using:

For \(f(x)=x^2+1\), the average gradient between \(x=1\) and \(x=3\) is:

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What CAPS Expects You to Know

The Grade 11 Functions knowledge statements this page is built from.

  1. 1

    Revise the effect of the parameters \(a\) and \(q\), and investigate and generalise the effect of the parameter \(p\), on the graphs of \(y=a(x+p)^2+q\), \(y=\dfrac{a}{x+p}+q\), and \(y=ab^{x+p}+q\) (\(b>0\), \(b\neq1\)).

  2. 2

    Investigate numerically the average gradient between two points on a curve, and develop an intuitive understanding of the concept of the gradient of a curve at a point.

How to Use This Lesson

A few practical notes before you start.

  • Make sure your Grade 10 \(a\) and \(q\) understanding is solid first — this lesson only adds \(p\) on top of it.
  • Every time you see \(x+p\), solve \(x+p=0\) rather than reading the sign off by eye — it's the single most common error at this level.
  • Average gradient uses the same formula as straight-line gradient. Don't overthink it: plug in \(f(x_1)\) and \(f(x_2)\).
  • Try each worked example yourself before pressing “Show solution.”

Functions & Graphs, Family by Family

Grade 11 adds the horizontal shift \(p\) to each graph family below, plus a first look at average gradient. Free, independent short videos — not made by Equation Station SA.

Parabola With p

\(y=a(x+p)^2+q\) — turning point \((-p,q)\). Solve \(x+p=0\) to find the shift.

Khan Academy · Shifting and scaling parabolas

Hyperbola With p

\(y=\dfrac{a}{x+p}+q\) — asymptotes \(x=-p\) and \(y=q\). The same general shift rule as every other family.

Khan Academy · Shifting functions introduction (general rule, applied here to the hyperbola)

Exponential With p

\(y=a\cdot b^{x+p}+q\) — horizontal asymptote \(y=q\), unaffected by \(p\).

Khan Academy · Transforming exponential graphs

Average Gradient

\(m=\dfrac{y_2-y_1}{x_2-x_1}\) between any two points on a curve — new this year, and a bridge to Grade 12 Calculus.

Khan Academy · Secant lines & average rate of change

Common Exam Mistakes

Avoid these errors. They cost marks every year.

Sign error on p

Reading \(x+p\) as if the shift were \(x=p\) instead of solving \(x+p=0\Rightarrow x=-p\).

Confusing p and q's effects

Attributing a vertical shift to \(p\), or a horizontal shift to \(q\) — they never swap roles.

Substituting into average gradient with only one full point

Both points need both coordinates before you can apply \(m=(y_2-y_1)/(x_2-x_1)\).

Doing combined transformations in the wrong order

"Shift then reflect" and "reflect then shift" give different equations — always follow the exact wording of the question.

This Topic in Other Grades

Functions & Graphs builds across all three FET grades — revise or preview the other levels here.

Grade 10

The four basic graph families, a and q only

Study this

Grade 12

Logarithmic graphs, inverse functions, full Paper 1 depth

Study this

Practise This Topic

You've done the notes above — now practise and test yourself.

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Past Papers
Functions & Graphs Grade 11 Past Papers

17 exam-style Grade 11 Functions questions arranged by cognitive level, with real citations from the DBE/provincial archive.

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Functions & Graphs Grade 11 Test Your Knowledge

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Frequently Asked Questions

Straight answers to common Grade 11 CAPS questions about the parameter p, average gradient, and what carries over from Grade 10.

What is new in Grade 11 Functions and Graphs?

Grade 11 introduces the horizontal shift parameter p on the parabola, hyperbola and exponential graphs, and gives learners their first introduction to average gradient between two points on a curve.

What does CAPS notation x+p mean for the turning point or asymptote?

CAPS Grade 11 notation writes the shift as x+p, so the turning point of a parabola or the vertical asymptote of a hyperbola sits at x=-p, not x=p. Forgetting to flip the sign is one of the most common Grade 11 exam errors.

Do I still need Grade 10 functions content?

Yes. The parameters a and q from Grade 10 are revised, not retaught, and Grade 11 exams test them cumulatively alongside the new parameter p.

What is average gradient?

Average gradient is the gradient of the straight line (chord) joining two points on a curve, calculated with the same formula as a straight-line gradient: m = (y2-y1)/(x2-x1).

Which resource should I open first for this topic?

Everything you need to learn the topic is on this page already. Once you've been through the notes above, work through the Past Question Papers, then finish with the Test Your Knowledge quiz as a self-check.