Draw the graphs of the quadratic polynomial f(x) = 3 - 2x – x2.

Solution

Let y = f(x) or, y = 3 - 2x – x2.

Let us list a few values of y = 3 – 2x – x2 corresponding to a few values of x as follows :

Thus, the following points lie on the graph of the polynomial y = 2 - 2x – x2:

(–5, –12), (–4, –5), (–3, 0), (–2, 4), (–1, 4), (0, 3), (1, 0), (2, –5),  (3, –12) and (4, –21).

Let plot these points on a graph paper and draw a smooth free hand curve passing through these points to obtain the graphs of y = 3 – 2x – x2. The curve thus obtained represents a parabola, as shown in figure. The highest point  P(-1, 4), is called a maximum points, is the vertex of the parabola. Vertical line through P is the axis of the parabola. Clearly, parabola is symmetric about the axis.

Observations :

Following observations from the graph of the polynomial f(x) = 3 – 2x – x2 is as follows:

(i) The coefficient of x2 in f(x) = 3 – 2x – x2  is - 1 i.e. a negative real number and so the parabola opens downwards.

(ii) D b2 – 4ax = 4 + 12 = 16 > 0. So, the parabola cuts x-axis two distinct points.

(iii) On comparing the polynomial 3 – 2x – x2 with ax2 + bc + c, we have a = –1, b = –2 and    c = 3. The vertex of the parabola is at the point (–1, 4) i.e. at , where             D = b2 – 4ac.