Question

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

Answer 1

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 :

x	-5	-4	-3	-2	-1	0	1	2	3	4
y=3-2x-x2	-12	-5	0	3	4	3	0	-5	-12	-21

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 (-b/2a,-D/4a), where D = b² - 4ac.

(iv) The polynomial f(x) = 3 - 2x - x² = (1 - x) (x + 3) is factorizable into two distinct linear factors (1 - x) and (x + 3). So, the parabola cuts X-axis at two distinct points (1, 0) and (-3, 0). The co-ordinates of these points are zeros of f(x).