Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2012 | June | Q#8

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Question

Where p and q are integers.

a.   Find the value of p and the value of q.

b.   Calculate the discriminant of  .

c.  On the axes on page 17, sketch the curve with equation  showing clearly the  coordinates of any points where the curve crosses the coordinate axes.

Solution

a.
 

We have the expression;

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

Next 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 ;

Therefore;

Therefore;

b.
 

For a quadratic equation , the expression for solution is;

Where  is called discriminant.

We are given;

Therefore;

c.    

We are required to sketch the curve with given equation;

We can rearrange  the given equation as;

It is evident that it is a quadratic equation.

To sketch a quadratic equation, a parabola, we need the coordinates of its vertex and x and y- intercepts, if any.

First we find the coordinates of vertex of this parabola.

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 downwards.

Vertex form of a quadratic equation is;

The given curve , can be written in this as demonstrated in (a);

Coordinates of the vertex are . Since this is a parabola opening downwards the vertex is the  maximum point on the graph. Here y-coordinate of vertex represents maximum value of  and x- coordinate of vertex represents corresponding value of .

For the given case, vertex is .

Next, we need x and y-intercepts of the parabola.

First we find the x-intercept of the parabola.

The point  at which curve (or line) intercepts x-axis, the value of . So we can find the  value of  coordinate by substituting  in the equation of the curve (or line).

Therefore, we substitute  in equation of the parabola.

Since this situation  is not possible, there is no solutions and , hence, no x-intercept of the  given parabola.

Next, we find the y-intercept of the parabola.

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, we substitute  in equation of the parabola.

Hence, coordinates of y-intercept of the parabola are .

We can sketch the parabola as shown below.

desmos-graph (10).png

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