# 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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**Question**

**a. **

** 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 .

**Solution**

**a.
**

** i.
**

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 ;

** ii.
**

It is evident that , hence, .

**b.
**

** i.
**

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 .

** ii.
**

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).

Therefore;

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

**c. **

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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