Graphs Flashcards
Reflection in the x-axis
When the (name of graphs) are reflected in the x-axis, the size and shape of the graphs both remain the same. However, the graphs are shifted vertically and their orientation changes (Eg. upside down). Therefore, the new (vertex/x-intercepts) of the parabola is at (coordinate(s))…(and so on for all the graphs and their key points and asymptotes)…
The new equations for these graphs are the same as their original ones, except the reflected graphs have a negative scalar in front of f(x) (eg. if the original equation is y = f(x), the new equation will be y = - f(x)), as they have been reflected in the x-axis and their orientation has changed.
The length and numbers of all the domains are also the same, as a reflection does not change the size of the shape, and a reflection in the x-axis keeps the same domain (eg. the new domains will be exactly the same as the original ones).
Therefore, the new equations will be… “ “ (equation) with the domain (domain) for the (graph) “ “
Reflection in the y-axis
When the (parabola, absolute value) are reflected in the y-axis, the size, shape and orientation of the graphs all remain the same. However, the graphs are shifted horizontally. Therefore, the new (vertex/x-intercepts) of the parabola is at (coordinate(s)) and the new vertex of the absolute value is at (coordinate).
When the (cubic, square root, hyperbola and exponential) are reflected in the y-axis, the size and shape of the graphs both remain the same. However, the graphs are shifted horizontally and their orientation changes. Therefore, the new (inflection point/x-intercepts) of the cubic is at (coordinate(s)), the new vertical asymptote of the hyperbola is at x = (__) and x = (__), and the new translated y-intercept of the exponential is at (coordinate).
The new equations for these graphs are the same as their original ones, except the reflected graphs have a negative scalar in front of x (eg. if the original equation is y = f(x), the new equation will be y = f(-x)), as they have been reflected in the y-axis and for (cubic, square root, hyperbola and exponential) their orientation has changed.
The length of all the domains are also the same, as a reflection does not change the size of the shape. However, the numbers will now be (positive/negative) as they are on the other side of the y-axis.
Therefore, the new equations will be… “ “ (equation) with the domain (domain) for the (graph) “ “
Horizontal translation
When the (name of graphs) are translated horizontally, the size, shape and orientation of the graphs all remain the same. However, the graphs are shifted horizontally. Therefore, the new (vertex/x-intercepts) of the parabola is at (coordinate(s))…(and so on for all the graphs and their key points and asymptotes)…
The new equations for these graphs are the same as their original ones, except the translated graphs (increase/decrease) the value of x by (amount) (eg. if the original equation is y = f(x), the new equation will be y = f(x±(amount))), as the graphs are shifted to the (left/right) by (amount) units.
The length of all the domains are also the same, as a translation does not change the size of a shape. However, the numbers are (increased/decreased) by (amount) units, as the position of the graphs are shifted horizontally by (amount) units.
Therefore, the new equations will be… “ “ (equation) with the domain (domain) for the (graph) “ “
Vertical translation
When the graphs are translated vertically, the size, shape and orientation of the graphs all remain the same. However, the graphs are shifted vertically. Therefore, the new (vertex/x-intercepts) of the parabola is at (coordinate(s))…(and so on for all the graphs and their key points and asymptotes)…
The new equations for these graphs are the same as their original ones, except the translated graphs (increase/decrease) the value of f(x) by (amount) (eg. if the original equation is y = f(x), the new equation will be y = f(x) ± (amount)), as the graphs are shifted (up/down) by (amount) units.
The length and numbers of all the domains are also the same, as a translation does not change the size of the shape, and a vertical translation keeps the same domain (eg. the new domains will be exactly the same as the original ones).
Therefore, the new equations will be… “ “ (equation) with the domain (domain) for the (graph) “ “
Parabola
The (part of image) can be modelled by a parabola which has (a vertex/x-intercepts) at (coordinate(s)) and goes through (coordinate).
- For the (one part of image), the equation must be (equation), thus (working to find value of a), the equation of (one part of image) is (equation), with the domain (domain).
Cubic
The (part of image) can be modelled by a cubic graph which has (an inflexion point/x-intercepts) at (coordinate(s) and goes through (coordinate).
- For the (one part of image), the equation must be (equation), thus (working to find value of a), the equation of (one part of image) is (equation), with the domain (domain).
Square root
I will use square root functions to model the (part of image), as they increase slowly and have starting points of (coordinate) (for (left/right, up/down) square root) and (coordinate) (for (left/right, up/down) square root).
- For the (left/right, up/down) square root,
- If it is a reflected graph… (it has been reflected, so)-
the equation must be (equation). The graph also goes through (coordinate), thus (working to find value of a), the equation of the (left/right, up/down) square root is (equation), with the domain (domain).
- “ “
Absolute value
I will use absolute value functions to model to the (part of image), as it is a straight line.
OR
I will use absolute value functions to model to the (part of image), as it is a symmetrical V-shape (meaning the gradient on either side is the negative of the other).
- For the (one part of image), the vertex is at (coordinate), and the gradient of the line(s) are (gradient), thus, the equation of (one part of image) is (equation), with the domain (domain).
- ” “
Hyperbola
I will use hyperbola to model the (part of image), as they both appear to have a horizontal asymptote at y = (amount), and vertical asymptote at x = (amount) (for (left/right, up/down) hyperbola), and x = (amount) (for (left/right, up/down) hyperbola).
- For the (left/right, up/down) hyperbola, the equation must be (equation). The graph also goes through (coordinate), thus (working to find value of a), the equation of the (left/right, up/down) hyperbola is (equation), with the domain (domain).
- ” “
Exponential
I will use exponential functions to model the (part of image), as they both appear to have a horizontal asymptote at y = (amount).
- For the (one part of image), the translated “y-intercept” is at (coordinate),
- If it is a reflected graph… (and it has been reflected, so)-
the equation must be (equation). The graph also goes through (coordinate), thus (working to find value of a), the equation of (one part of image) is (equation), with the domain (domain).
- “ “
- NOTE - if the exponential graph, is reflected in the x-axis, the horizontal asymptote which changes in the graph (eg. reflected in the x-axis) will need to be changed in the equation also (in addition to adding a negative scalar in front of f(x)).
Eg.
From y = 3^(4 - x)^ + 1
to y = -3^(4 - x)^ - 1
How to find the gradient of an absolute value line
Find the Rise/Run of the positive line
Equations for parabola
x-intercepts -
y = ±a (x±b)(x±c)
vertex -
y = ±a (x±b)^2 ±c
Equations for cubic
x-intercepts -
y = ±a (x±b)(x±c)(x±d)
OR
y = ±a (x±b)(x±c)^2 (if it only ‘touches the x-axis)
inflection point -
y = ±a (x±b)^3 ±c
Equations for square root
vertex -
y = ±a √(x±b) ±c
OR
y = ±a √-(x±b) ±c (if it is reflected/a decay curve)
Equations for absolute value
vertex -
y = ±a | x±b | ±c (where a = the gradient of the lines)
