{"id":100,"date":"2026-08-07T11:33:07","date_gmt":"2026-08-07T11:33:07","guid":{"rendered":"https:\/\/sqoolpapers.co.za\/notes\/functions-and-graphs-g12-maths-t4p1-test-3\/"},"modified":"2026-08-22T16:31:04","modified_gmt":"2026-08-22T16:31:04","slug":"functions-and-graphs-g12-maths-t4p1","status":"publish","type":"post","link":"https:\/\/repstaq.com\/study-guide-docs\/functions-and-graphs-g12-maths-t4p1\/","title":{"rendered":"Functions and Graphs-G12-Maths-T4P1"},"content":{"rendered":"\n

Functions and Graphs<\/h2>\n\n\n\n

Functions describe relationships between variables. A graph represents this relationship on the Cartesian plane. In examinations, you must recognise the type of function, determine important features, sketch graphs accurately and interpret intersections or intervals.<\/p>\n\n\n\n

Linear functions<\/h2>\n\n\n\n

The general form of a linear function is:<\/p>\n\n\n\n

f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>m<\/mi>x<\/mi>+<\/mo>c<\/mi><\/mrow>f(x)=mx+c<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

The constant m is the gradient, and c is the y-intercept. The gradient measures the change in the y-values compared with the change in the x-values:<\/p>\n\n\n\n

m<\/mi>=<\/mo>y<\/mi>2<\/mn><\/msub>\u2212<\/mo>y<\/mi>1<\/mn><\/msub><\/mrow>x<\/mi>2<\/mn><\/msub>\u2212<\/mo>x<\/mi>1<\/mn><\/msub><\/mrow><\/mfrac><\/mrow>m=\\frac{y_2-y_1}{x_2-x_1}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

If the gradient is positive, the graph increases from left to right. If it is negative, the graph decreases. If it is zero, the graph is horizontal.<\/p>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

Determine the equation of the straight line passing through the points with coordinates:<\/p>\n\n\n\n

A<\/mi>(<\/mo>1<\/mn>;<\/mo>3<\/mn>)<\/mo><\/mspace>and<\/mtext><\/mspace>B<\/mi>(<\/mo>4<\/mn>;<\/mo>9<\/mn>)<\/mo><\/mrow>A(1;3)\\quad\\text{and}\\quad B(4;9)<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

First calculate the gradient:<\/p>\n\n\n\n

m<\/mi>=<\/mo>9<\/mn>\u2212<\/mo>3<\/mn><\/mrow>4<\/mn>\u2212<\/mo>1<\/mn><\/mrow><\/mfrac><\/mrow>m=\\frac{9-3}{4-1}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
m<\/mi>=<\/mo>6<\/mn>3<\/mn><\/mfrac>=<\/mo>2<\/mn><\/mrow>m=\\frac{6}{3}=2<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Substitute one point into the general equation:<\/p>\n\n\n\n

y<\/mi>=<\/mo>m<\/mi>x<\/mi>+<\/mo>c<\/mi><\/mrow>y=mx+c<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
3<\/mn>=<\/mo>2<\/mn>(<\/mo>1<\/mn>)<\/mo>+<\/mo>c<\/mi><\/mrow>3=2(1)+c<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
c<\/mi>=<\/mo>1<\/mn><\/mrow>c=1<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Therefore:<\/p>\n\n\n\n

f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>2<\/mn>x<\/mi>+<\/mo>1<\/mn><\/mrow>f(x)=2x+1<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

To calculate the x-intercept, let the y-value equal zero:<\/p>\n\n\n\n

0<\/mn>=<\/mo>2<\/mn>x<\/mi>+<\/mo>1<\/mn><\/mrow>0=2x+1<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
x<\/mi>=<\/mo>\u2212<\/mo>1<\/mn>2<\/mn><\/mfrac><\/mrow>x=-\\frac{1}{2}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Quadratic functions<\/h2>\n\n\n\n

The general form of a quadratic function is:<\/p>\n\n\n\n

f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>a<\/mi>x<\/mi>2<\/mn><\/msup>+<\/mo>b<\/mi>x<\/mi>+<\/mo>c<\/mi><\/mrow>f(x)=ax^2+bx+c<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Its graph is a parabola. If the value of a is positive, the parabola opens upwards and has a minimum turning point. If the value of a is negative, it opens downwards and has a maximum turning point.<\/p>\n\n\n\n

The axis of symmetry is:<\/p>\n\n\n\n

x<\/mi>=<\/mo>\u2212<\/mo>b<\/mi>2<\/mn>a<\/mi><\/mrow><\/mfrac><\/mrow>x=-\\frac{b}{2a}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

Determine the intercepts and turning point of:<\/p>\n\n\n\n

f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>x<\/mi>2<\/mn><\/msup>\u2212<\/mo>4<\/mn>x<\/mi>\u2212<\/mo>5<\/mn><\/mrow>f(x)=x^2-4x-5<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

For the y-intercept, let the x-value equal zero:<\/p>\n\n\n\n

f<\/mi>(<\/mo>0<\/mn>)<\/mo>=<\/mo>0<\/mn>2<\/mn><\/msup>\u2212<\/mo>4<\/mn>(<\/mo>0<\/mn>)<\/mo>\u2212<\/mo>5<\/mn><\/mrow>f(0)=0^2-4(0)-5<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
f<\/mi>(<\/mo>0<\/mn>)<\/mo>=<\/mo>\u2212<\/mo>5<\/mn><\/mrow>f(0)=-5<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

The y-intercept is therefore:<\/p>\n\n\n\n

(<\/mo>0<\/mn>;<\/mo>\u2212<\/mo>5<\/mn>)<\/mo><\/mrow>(0;-5)<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

For the x-intercepts, let the function equal zero:<\/p>\n\n\n\n

x<\/mi>2<\/mn><\/msup>\u2212<\/mo>4<\/mn>x<\/mi>\u2212<\/mo>5<\/mn>=<\/mo>0<\/mn><\/mrow>x^2-4x-5=0<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
(<\/mo>x<\/mi>\u2212<\/mo>5<\/mn>)<\/mo>(<\/mo>x<\/mi>+<\/mo>1<\/mn>)<\/mo>=<\/mo>0<\/mn><\/mrow>(x-5)(x+1)=0<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
x<\/mi>=<\/mo>5<\/mn><\/mspace>or<\/mtext><\/mspace>x<\/mi>=<\/mo>\u2212<\/mo>1<\/mn><\/mrow>x=5\\quad\\text{or}\\quad x=-1<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

The x-intercepts are:<\/p>\n\n\n\n

(<\/mo>5<\/mn>;<\/mo>0<\/mn>)<\/mo><\/mspace>and<\/mtext><\/mspace>(<\/mo>\u2212<\/mo>1<\/mn>;<\/mo>0<\/mn>)<\/mo><\/mrow>(5;0)\\quad\\text{and}\\quad(-1;0)<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Calculate the x-coordinate of the turning point:<\/p>\n\n\n\n

x<\/mi>=<\/mo>\u2212<\/mo>\u2212<\/mo>4<\/mn><\/mrow>2<\/mn>(<\/mo>1<\/mn>)<\/mo><\/mrow><\/mfrac><\/mrow>x=-\\frac{-4}{2(1)}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
x<\/mi>=<\/mo>2<\/mn><\/mrow>x=2<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Calculate the corresponding y-coordinate:<\/p>\n\n\n\n

f<\/mi>(<\/mo>2<\/mn>)<\/mo>=<\/mo>2<\/mn>2<\/mn><\/msup>\u2212<\/mo>4<\/mn>(<\/mo>2<\/mn>)<\/mo>\u2212<\/mo>5<\/mn><\/mrow>f(2)=2^2-4(2)-5<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
f<\/mi>(<\/mo>2<\/mn>)<\/mo>=<\/mo>\u2212<\/mo>9<\/mn><\/mrow>f(2)=-9<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

The turning point is:<\/p>\n\n\n\n

(<\/mo>2<\/mn>;<\/mo>\u2212<\/mo>9<\/mn>)<\/mo><\/mrow>(2;-9)<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
\"\"\/<\/a><\/figure>\n\n\n\n

Hyperbolic functions<\/h2>\n\n\n\n

The standard form of a hyperbolic function is:<\/p>\n\n\n\n

f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>a<\/mi>x<\/mi>\u2212<\/mo>p<\/mi><\/mrow><\/mfrac>+<\/mo>q<\/mi><\/mrow>f(x)=\\frac{a}{x-p}+q<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

The graph has two separate branches. Its asymptotes are:<\/p>\n\n\n\n

x<\/mi>=<\/mo>p<\/mi><\/mrow>x=p<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
y<\/mi>=<\/mo>q<\/mi><\/mrow>y=q<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

The point where the asymptotes intersect is the centre of the hyperbola:<\/p>\n\n\n\n

(<\/mo>p<\/mi>;<\/mo>q<\/mi>)<\/mo><\/mrow>(p;q)<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

If a is positive, the branches lie above-right and below-left relative to the centre. If a is negative, the branches lie above-left and below-right.<\/p>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

Consider the function:<\/p>\n\n\n\n

f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>2<\/mn>x<\/mi>\u2212<\/mo>2<\/mn><\/mrow><\/mfrac>\u2212<\/mo>1<\/mn><\/mrow>f(x)=\\frac{2}{x-2}-1<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Its asymptotes are:<\/p>\n\n\n\n

x<\/mi>=<\/mo>2<\/mn><\/mrow>x=2<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
y<\/mi>=<\/mo>\u2212<\/mo>1<\/mn><\/mrow>y=-1<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

To calculate the x-intercept:<\/p>\n\n\n\n

0<\/mn>=<\/mo>2<\/mn>x<\/mi>\u2212<\/mo>2<\/mn><\/mrow><\/mfrac>\u2212<\/mo>1<\/mn><\/mrow>0=\\frac{2}{x-2}-1<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
1<\/mn>=<\/mo>2<\/mn>x<\/mi>\u2212<\/mo>2<\/mn><\/mrow><\/mfrac><\/mrow>1=\\frac{2}{x-2}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
x<\/mi>\u2212<\/mo>2<\/mn>=<\/mo>2<\/mn><\/mrow>x-2=2<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
x<\/mi>=<\/mo>4<\/mn><\/mrow>x=4<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

To calculate the y-intercept:<\/p>\n\n\n\n

f<\/mi>(<\/mo>0<\/mn>)<\/mo>=<\/mo>2<\/mn>0<\/mn>\u2212<\/mo>2<\/mn><\/mrow><\/mfrac>\u2212<\/mo>1<\/mn><\/mrow>f(0)=\\frac{2}{0-2}-1<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
f<\/mi>(<\/mo>0<\/mn>)<\/mo>=<\/mo>\u2212<\/mo>2<\/mn><\/mrow>f(0)=-2<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

An examination derivation uses the asymptotes and points on the graph to determine the parameters before obtaining a function of this form. It also identifies the related graph transformation as a translation downwards or to the right. <\/p>\n\n\n\n

Exponential functions<\/h2>\n\n\n\n

The general exponential function is:<\/p>\n\n\n\n

f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>a<\/mi>b<\/mi>x<\/mi><\/msup>+<\/mo>q<\/mi><\/mrow>f(x)=ab^x+q<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

The base must satisfy:<\/p>\n\n\n\n

b<\/mi>><\/mo>0<\/mn><\/mspace>and<\/mtext><\/mspace>b<\/mi>\u2260<\/mo>1<\/mn><\/mrow>b>0\\quad\\text{and}\\quad b\\ne1<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

The horizontal asymptote is:<\/p>\n\n\n\n

y<\/mi>=<\/mo>q<\/mi><\/mrow>y=q<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

If the base is greater than one, the graph increases. If the base lies between zero and one, the graph decreases.<\/p>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

Consider:<\/p>\n\n\n\n

f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>2<\/mn>x<\/mi><\/msup>\u2212<\/mo>4<\/mn><\/mrow>f(x)=2^x-4<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

The horizontal asymptote is:<\/p>\n\n\n\n

y<\/mi>=<\/mo>\u2212<\/mo>4<\/mn><\/mrow>y=-4<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

For the y-intercept:<\/p>\n\n\n\n

f<\/mi>(<\/mo>0<\/mn>)<\/mo>=<\/mo>2<\/mn>0<\/mn><\/msup>\u2212<\/mo>4<\/mn><\/mrow>f(0)=2^0-4<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
f<\/mi>(<\/mo>0<\/mn>)<\/mo>=<\/mo>1<\/mn>\u2212<\/mo>4<\/mn>=<\/mo>\u2212<\/mo>3<\/mn><\/mrow>f(0)=1-4=-3<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

For the x-intercept:<\/p>\n\n\n\n

0<\/mn>=<\/mo>2<\/mn>x<\/mi><\/msup>\u2212<\/mo>4<\/mn><\/mrow>0=2^x-4<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
2<\/mn>x<\/mi><\/msup>=<\/mo>4<\/mn><\/mrow>2^x=4<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
2<\/mn>x<\/mi><\/msup>=<\/mo>2<\/mn>2<\/mn><\/msup><\/mrow>2^x=2^2<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
x<\/mi>=<\/mo>2<\/mn><\/mrow>x=2<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Inverse functions<\/h2>\n\n\n\n

An inverse function reverses the operation of the original function. To determine an inverse, replace the function notation with the y-variable, interchange the x-variable and y-variable, and make the y-variable the subject.<\/p>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

Determine the inverse of:<\/p>\n\n\n\n

f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>2<\/mn>x<\/mi>+<\/mo>6<\/mn><\/mrow>f(x)=2x+6<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
y<\/mi>=<\/mo>2<\/mn>x<\/mi>+<\/mo>6<\/mn><\/mrow>y=2x+6<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Interchange the variables:<\/p>\n\n\n\n

x<\/mi>=<\/mo>2<\/mn>y<\/mi>+<\/mo>6<\/mn><\/mrow>x=2y+6<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Make the y-variable the subject:<\/p>\n\n\n\n

x<\/mi>\u2212<\/mo>6<\/mn>=<\/mo>2<\/mn>y<\/mi><\/mrow>x-6=2y<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
y<\/mi>=<\/mo>x<\/mi>\u2212<\/mo>6<\/mn><\/mrow>2<\/mn><\/mfrac><\/mrow>y=\\frac{x-6}{2}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Therefore:<\/p>\n\n\n\n

f<\/mi>\u2212<\/mo>1<\/mn><\/mrow><\/msup>(<\/mo>x<\/mi>)<\/mo>=<\/mo>x<\/mi>\u2212<\/mo>6<\/mn><\/mrow>2<\/mn><\/mfrac><\/mrow>f^{-1}(x)=\\frac{x-6}{2}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

A function and its inverse are reflections of each other in the line:<\/p>\n\n\n\n

y<\/mi>=<\/mo>x<\/mi><\/mrow>y=x<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

The domain of the original function becomes the range of its inverse, while its range becomes the domain of the inverse. A quadratic function must usually have a restricted domain before its inverse is also a function.<\/p>\n\n\n\n

Domain and range<\/h2>\n\n\n\n

The domain is the set of permissible x-values. The range is the set of possible y-values.<\/p>\n\n\n\n

For a linear function with a non-zero gradient:<\/p>\n\n\n\n

x<\/mi>\u2208<\/mo>\u211d<\/mi><\/mrow>x\\in\\mathbb{R}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
y<\/mi>\u2208<\/mo>\u211d<\/mi><\/mrow>y\\in\\mathbb{R}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

For a quadratic function, the domain is all real numbers. Its range depends on the turning point. For the earlier quadratic example with minimum value negative nine:<\/p>\n\n\n\n

x<\/mi>\u2208<\/mo>\u211d<\/mi><\/mrow>x\\in\\mathbb{R}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
y<\/mi>\u2265<\/mo>\u2212<\/mo>9<\/mn><\/mrow>y\\geq-9<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

For the hyperbola:<\/p>\n\n\n\n

f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>2<\/mn>x<\/mi>\u2212<\/mo>2<\/mn><\/mrow><\/mfrac>\u2212<\/mo>1<\/mn><\/mrow>f(x)=\\frac{2}{x-2}-1<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

the domain and range are:<\/p>\n\n\n\n

x<\/mi>\u2208<\/mo>\u211d<\/mi>,<\/mo><\/mspace>x<\/mi>\u2260<\/mo>2<\/mn><\/mrow>x\\in\\mathbb{R},\\quad x\\ne2<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
y<\/mi>\u2208<\/mo>\u211d<\/mi>,<\/mo><\/mspace>y<\/mi>\u2260<\/mo>\u2212<\/mo>1<\/mn><\/mrow>y\\in\\mathbb{R},\\quad y\\ne-1<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

For the exponential function:<\/p>\n\n\n\n

f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>2<\/mn>x<\/mi><\/msup>\u2212<\/mo>4<\/mn><\/mrow>f(x)=2^x-4<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

the domain and range are:<\/p>\n\n\n\n

x<\/mi>\u2208<\/mo>\u211d<\/mi><\/mrow>x\\in\\mathbb{R}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
y<\/mi>><\/mo>\u2212<\/mo>4<\/mn><\/mrow>y>-4<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Intercepts and turning points<\/h2>\n\n\n\n

An x-intercept occurs where a graph crosses or touches the x-axis. Set the function equal to zero.<\/p>\n\n\n\n

f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>0<\/mn><\/mrow>f(x)=0<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

A y-intercept occurs where a graph crosses the y-axis. Substitute zero for the x-value.<\/p>\n\n\n\n

f<\/mi>(<\/mo>0<\/mn>)<\/mo>=<\/mo>y<\/mi><\/mrow>f(0)=y<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

A quadratic function has one turning point. It may also have two, one or no real x-intercepts. The discriminant can be used to determine this:<\/p>\n\n\n\n

\u0394<\/mi><\/mrow>=<\/mo>b<\/mi>2<\/mn><\/msup>\u2212<\/mo>4<\/mn>a<\/mi>c<\/mi><\/mrow>\\Delta=b^2-4ac<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
\u0394<\/mi><\/mrow>><\/mo>0<\/mn><\/mrow>\\Delta>0<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

This means that there are two distinct x-intercepts.<\/p>\n\n\n\n

\u0394<\/mi><\/mrow>=<\/mo>0<\/mn><\/mrow>\\Delta=0<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

This means that there is one repeated x-intercept.<\/p>\n\n\n\n

\u0394<\/mi><\/mrow><<\/mo>0<\/mn><\/mrow>\\Delta<0<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

This means that there are no real x-intercepts.<\/p>\n\n\n\n

Asymptotes<\/h2>\n\n\n\n

An asymptote is a line that a graph approaches but does not reach under normal conditions.<\/p>\n\n\n\n

For a hyperbola:<\/p>\n\n\n\n

f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>a<\/mi>x<\/mi>\u2212<\/mo>p<\/mi><\/mrow><\/mfrac>+<\/mo>q<\/mi><\/mrow>f(x)=\\frac{a}{x-p}+q<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

the vertical and horizontal asymptotes are:<\/p>\n\n\n\n

x<\/mi>=<\/mo>p<\/mi><\/mrow>x=p<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
y<\/mi>=<\/mo>q<\/mi><\/mrow>y=q<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

For an exponential function:<\/p>\n\n\n\n

f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>a<\/mi>b<\/mi>x<\/mi><\/msup>+<\/mo>q<\/mi><\/mrow>f(x)=ab^x+q<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

the horizontal asymptote is:<\/p>\n\n\n\n

y<\/mi>=<\/mo>q<\/mi><\/mrow>y=q<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Asymptotes are normally drawn as broken lines and labelled clearly. They help determine the domain, range and overall position of the graph.<\/p>\n\n\n\n

Transformations of graphs<\/h2>\n\n\n\n

Starting with a basic graph, transformations change its position or shape.<\/p>\n\n\n\n

A vertical translation is represented by:<\/p>\n\n\n\n

y<\/mi>=<\/mo>f<\/mi>(<\/mo>x<\/mi>)<\/mo>+<\/mo>q<\/mi><\/mrow>y=f(x)+q<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

If q is positive, the graph moves upwards. If q is negative, it moves downwards.<\/p>\n\n\n\n

A horizontal translation is represented by:<\/p>\n\n\n\n

y<\/mi>=<\/mo>f<\/mi>(<\/mo>x<\/mi>\u2212<\/mo>p<\/mi>)<\/mo><\/mrow>y=f(x-p)<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

If p is positive, the graph moves to the right. If p is negative, it moves to the left.<\/p>\n\n\n\n

Reflection in the x-axis is represented by:<\/p>\n\n\n\n

y<\/mi>=<\/mo>\u2212<\/mo>f<\/mi>(<\/mo>x<\/mi>)<\/mo><\/mrow>y=-f(x)<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Reflection in the y-axis is represented by:<\/p>\n\n\n\n

y<\/mi>=<\/mo>f<\/mi>(<\/mo>\u2212<\/mo>x<\/mi>)<\/mo><\/mrow>y=f(-x)<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

A change from one related hyperbola to another may be described as a vertical or horizontal translation, provided the direction and number of units are stated correctly. <\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Graph interpretation<\/h2>\n\n\n\n

When interpreting a graph, read coordinates and intervals carefully.<\/p>\n\n\n\n

A graph is increasing where the y-values increase as the x-values increase. It is decreasing where the y-values decrease as the x-values increase.<\/p>\n\n\n\n

The function is positive where the graph lies above the x-axis:<\/p>\n\n\n\n

f<\/mi>(<\/mo>x<\/mi>)<\/mo>><\/mo>0<\/mn><\/mrow>f(x)>0<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

The function is negative where the graph lies below the x-axis:<\/p>\n\n\n\n

f<\/mi>(<\/mo>x<\/mi>)<\/mo><<\/mo>0<\/mn><\/mrow>f(x)<0<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

To determine where one graph lies above another, identify the intersections and compare the graphs between these points:<\/p>\n\n\n\n

f<\/mi>(<\/mo>x<\/mi>)<\/mo>><\/mo>g<\/mi>(<\/mo>x<\/mi>)<\/mo><\/mrow>f(x)>g(x)<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Remember to use strict inequalities if the points of intersection are excluded and inclusive inequalities if they are included.<\/p>\n\n\n\n

Solving equations using graphs<\/h2>\n\n\n\n

The solutions of an equation involving two functions are the x-coordinates of their points of intersection.<\/p>\n\n\n\n

For example, to solve:<\/p>\n\n\n\n

x<\/mi>2<\/mn><\/msup>\u2212<\/mo>4<\/mn>x<\/mi>\u2212<\/mo>5<\/mn>=<\/mo>2<\/mn>x<\/mi>+<\/mo>1<\/mn><\/mrow>x^2-4x-5=2x+1<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

draw or consider the graphs:<\/p>\n\n\n\n

f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>x<\/mi>2<\/mn><\/msup>\u2212<\/mo>4<\/mn>x<\/mi>\u2212<\/mo>5<\/mn><\/mrow>f(x)=x^2-4x-5<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
g<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>2<\/mn>x<\/mi>+<\/mo>1<\/mn><\/mrow>g(x)=2x+1<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Algebraically, the intersection points are found as follows:<\/p>\n\n\n\n

x<\/mi>2<\/mn><\/msup>\u2212<\/mo>4<\/mn>x<\/mi>\u2212<\/mo>5<\/mn>=<\/mo>2<\/mn>x<\/mi>+<\/mo>1<\/mn><\/mrow>x^2-4x-5=2x+1<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
x<\/mi>2<\/mn><\/msup>\u2212<\/mo>6<\/mn>x<\/mi>\u2212<\/mo>6<\/mn>=<\/mo>0<\/mn><\/mrow>x^2-6x-6=0<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Using the quadratic formula:<\/p>\n\n\n\n

x<\/mi>=<\/mo>\u2212<\/mo>b<\/mi>\u00b1<\/mo>b<\/mi>2<\/mn><\/msup>\u2212<\/mo>4<\/mn>a<\/mi>c<\/mi><\/mrow><\/msqrt><\/mrow>2<\/mn>a<\/mi><\/mrow><\/mfrac><\/mrow>x=\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
x<\/mi>=<\/mo>6<\/mn>\u00b1<\/mo>(<\/mo>\u2212<\/mo>6<\/mn>)<\/mo>2<\/mn><\/msup>\u2212<\/mo>4<\/mn>(<\/mo>1<\/mn>)<\/mo>(<\/mo>\u2212<\/mo>6<\/mn>)<\/mo><\/mrow><\/msqrt><\/mrow>2<\/mn>(<\/mo>1<\/mn>)<\/mo><\/mrow><\/mfrac><\/mrow>x=\\frac{6\\pm\\sqrt{(-6)^2-4(1)(-6)}}{2(1)}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
x<\/mi>=<\/mo>6<\/mn>\u00b1<\/mo>60<\/mn><\/msqrt><\/mrow>2<\/mn><\/mfrac><\/mrow>x=\\frac{6\\pm\\sqrt{60}}{2}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
x<\/mi>=<\/mo>3<\/mn>\u00b1<\/mo>15<\/mn><\/msqrt><\/mrow>x=3\\pm\\sqrt{15}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

On a graph, these solutions are read from the x-coordinates where the straight line intersects the parabola. Simultaneous equations can similarly be solved by substitution or by reading the intersection coordinates. Complete examination working should show the equations, substitution, standard form and both coordinate values. <\/p>\n\n\n\n

Exam Tip<\/strong><\/p>\n\n\n\n

Always label the axes, intercepts, turning points and asymptotes. Show substitutions and important algebraic steps. When giving an interval, check whether the endpoints must be included. Use the shape, domain, range and asymptotes to confirm that your calculated graph is reasonable.<\/p>\n","protected":false},"excerpt":{"rendered":"

Functions and Graphs Functions describe relationships between variables. A graph represents this relationship on the Cartesian plane. In examinations, you must recognise the type of function, determine important features, sketch graphs accurately and interpret intersections or intervals. Linear functions The general form of a linear function is: The constant m is the gradient, and c […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"_breakdance_hide_in_design_set":false,"_breakdance_tags":"","footnotes":""},"categories":[1],"tags":[],"class_list":["post-100","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"acf":{"document_name":"Functions and Graphs","grade":12,"subject":"Mathematics","term":4,"paper":1},"_links":{"self":[{"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/posts\/100","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/comments?post=100"}],"version-history":[{"count":4,"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/posts\/100\/revisions"}],"predecessor-version":[{"id":235,"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/posts\/100\/revisions\/235"}],"wp:attachment":[{"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/media?parent=100"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/categories?post=100"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/tags?post=100"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}