# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2017 | June | Q#5

Question

f(x) = x2 – 8x + 19

a.   (a) Express f(x) in the form (x + a)2 + b, where a and b are constants.

The curve C with equation y = f(x) crosses the y-axis at the point P and has a minimum point at the  point Q.

b.   Sketch the graph of C showing the coordinates of point P and the coordinates of point Q.

c.   Find the distance PQ, writing your answer as a simplified surd.

Solution

a)

We have 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 ;

b)

It is evident that  is a quadratic equation. Therefore, we are required to sketch  the parabola with coordinates of y-intercept and vertex (minimum point).

It is evident that curve is a parabola (quadratic equation ie polynomial of degree 2).

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;

As demonstrated in (a), we can write the given equation of parabola as;

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  is not possible, we can deduce that there are no x-intercepts of the 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.

c)

We are required to find the distance PQ.

Expression for the distance between two given points  and is:

We have found coordinates of both point P and Q in (b) as (0,19) and (4,3).

Therefore;

Since ;