Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | 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, x-intercept(s) and y-intercept.

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 y-coordinate of vertex represents least value of
 and x-coordinate of vertex represents  corresponding value of .

For the given case, vertex is .

Next we find coordinates of x-intercept(s). Since vertex lies on x-axis, there will be no x-intercept point other than vertex.

Finally we need to find coordinates of y-intercept.

The point  at which curve (or line) intercepts y-axis, 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 y-intercept are (0,9).

We can sketch the graph of  as shown below.

Untitled.png

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
positive x-direction

 units in
positive y-direction

Coordinates

Function

 units in
negative x-direction

 units in
positive y-direction

Coordinates

Function

 units in
positive x-direction

 units in
negative y-direction

Coordinates

Function

 units in
negative x-direction

 units in
negative y-direction

Coordinates

Function

 units in
positive y-direction

Coordinates

Function

 units in
negative y-direction

Coordinates

Function

 units in
positive x-direction

Coordinates

Function

 units in
negative x-direction

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 y-axis.

Translation through vector  represents the move,  units in the x-direction and  units in the  positive y-direction.

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 y-axis.

It is also evident from the above table that only y-coordinates of the graph change whereas x-  coordinates of the graph will remain unchanged.  

Hence, the new function has all the x-coordinates same as that of original given function  whereas all the y-coordinates are k units more than that of original given function.  

It is shown in the figure below.

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