Past Papers’ Solutions  Edexcel  AS & A level  Mathematics  Core Mathematics 1 (C16663/01)  Year 2006  June  Q#3
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Question
On separate diagrams, sketch the graphs of
a. y = (x+3)^{2}
b. y = (x+3)^{2}+k, where k is a positive constant.
Show on each sketch the coordinates of each point at which the graph meets the axes.
Solution
a.
We are required to sketch the graph of;
It is evident that it is a quadratic equation and, hence, its graph will be a parabola.
We can sketch the parabola by finding coordinates of vertex, xintercept(s) and yintercept.
First we find coordinates of vertex.
Standard form of quadratic equation is;
The graph of quadratic equation is a parabola. If (‘a’ is positive) then parabola opens upwards and its vertex is the minimum point on the graph.
If (‘a’ is negative) then parabola opens downwards and its vertex is the maximum point on the graph.
Vertex form of a quadratic equation is;
The given curve , is in vertex form.
We can write it in standard form as;
Coordinates of the vertex are . Since this is a parabola opening upwards the vertex is the minimum point on the graph.
Here ycoordinate of vertex represents least value of and xcoordinate of vertex represents corresponding value of .
For the given case, vertex is .
Next we find coordinates of xintercept(s). Since vertex lies on xaxis, there will be no xintercept point other than vertex.
Finally we need to find coordinates of yintercept.
The point at which curve (or line) intercepts yaxis, the value of . So we can find the value of coordinate by substituting in the equation of the curve (or line).
Therefore, substituting ;
Hence, coordinates of yintercept are (0,9).
We can sketch the graph of as shown below.
b.
We are required to sketch the graph of;
We know that represents ‘translation’ in transformation of given function. Hence, and .
But we have sketched the graph of in (a), therefore, the same sketch can be translated to sketch the graph of .
For this purpose, and .
Original 
Transformed 
Translation Vector 
Movement 

Function 



units in units in 
Coordinates 



Function 



units in units in 
Coordinates 



Function 



units in units in 
Coordinates 



Function 



units in units in 
Coordinates 



Function 



units in 
Coordinates 



Function 



units in 
Coordinates 



Function 



units in 
Coordinates 



Function 



units in 
Coordinates 


From the above table, as highlighted, it is evident that we are required to transform the function into , where , (since k is a positive constant) therefore it is case of translation in the direction of positive yaxis.
Translation through vector represents the move, units in the xdirection and units in the positive ydirection.
Translation through vector transforms the function into or .
Transformation of the function into or results from translation through vector .
Translation vector transforms the function into or which means shift upwards along yaxis.
It is also evident from the above table that only ycoordinates of the graph change whereas x coordinates of the graph will remain unchanged.
Hence, the new function has all the xcoordinates same as that of original given function whereas all the ycoordinates are k units more than that of original given function.
It is shown in the figure below.
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