Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2011 | June | Q#4
a. Express in the form , where p and q are rational numbers.
b. A curve has equation .
i. Find the coordinates of the vertex of the curve.
ii. State the equation of the line of symmetry of the curve.
iii. Sketch the curve, stating the value of the intercept on the y-axis.
iv. 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.
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 ;
We are given equation of the curve as;
We are required to find the coordinates of the vertex of the curve.
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) 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 curve coordinates of vertex are .
Line of symmetry of a parabola is the vertical line passing through its vertex.
Therefore, for the given curve vertex is and equation of line of symmetry is;
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.
We can map the graph of onto the graph as follows.
We have shown in (a) that equation of given curve can be written as;
Therefore, We need to map the graph of onto the graph .
In general, a translation of transforms the graph of into the graph of .
Translation vector represents the move, units in the positive x-direction then units in the positive y-direction.
Therefore, for the given case, translation of by vector maps the graph onto .