Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2009 | January | Q#4

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            i.               Express  in the form  , where p and q are integers.

          ii.               Hence show that  is always positive.

b.   A curve has equation  .

                           i.               Write down the coordinates of the minimum point of the curve.

                         ii.               Sketch the curve, showing the value of the intercept on the y-axis.

c.   Describe the geometrical transformation that maps the graph of  onto the graph of .




We have the expression;

We use method of “completing square” to obtain the desired form.

We complete the square for the terms which involve .

We have the algebraic formula;

For the given case we can compare the given terms with the formula as below;

Therefore we can deduce that;

Hence we can write;

To complete the square we can add and subtract the deduced value of ;


It is evident that , hence, .




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.

We recognize that given curve , is a parabola opening upwards.

Vertex form of a quadratic equation is;

The given curve , as demonstrated in (a:i) can be written in vertex 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 .

Since given curve is a parabola opening upwards, the vertex is the minimum point .



We are required to sketch  .

It is evident that it is quadratic equation which will form a parabola. To sketch a parabola we need  coordinates of vertex point and coordinates of x-intercepts.

From (b:i) we know that the vertex of the given parabola is a minimum point and has coordinates .

Since minimum point of upwards opening parabola is above x-axis there will be no x-intercepts. However, we need to check for 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).


Now we can sketch graph of given equation using this information.

desmos-graph (8).png


Moving a curve without altering its shape is called a translation.

We have;

We transform it into desired form;

The given curve , as demonstrated in (a:i), can be written in vertex form as;

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

In general, a translation of  transforms the graph of  into the graph of .

Therefore, translation vector of the given and desired graphs is .

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