Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2013 | January | Q#10
Question
a. Find the values of the constants a, b and c.
b. On the axes on page 27, sketch the curve with equation y = 4x2 + 8x + 3, showing clearly the coordinates of any points where the curve crosses the coordinate axes.
Solution
a.
We are given that;
In order to find the values of the constants a, b and c, we need to write the left hand side of the equation in the same form as right hand side of the equation.
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 ;
Hence;
Therefore;
b.
We are required to sketch the curve with given equation;
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 upwards.
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.
Now we have two options.
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Therefore, there are two x-intercepts of the given parabolic curve with coordinates and
.
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.
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