Draw to graph for the said equation.
| x | 0 | 1 | 2 | 3 | 4 | -1 | -2 | -3 |
| y | -4 | -6 | -6 | -4 | 0 | 0 | 6 | -14 |
For any quadratic polynomial $ax^2+bx+c$ , $a ≠ 0$, the graph of the corresponding equation $y = ax^2 + bx + c$ has one of the two shapes either open upwards like or open downwards like depending on whether $a > 0$ or $a < 0$. (These curves are called parabolas.)
Solving the equation by factorising the middle term; $x^2-3x-4$ =$(x^2+x-4x-4)$ =$x(x+1)-4(x+1)$
$x(x+1)-4(x+1) =0$ i.e $(x+1)(x-4)=0$
i.e either $x+1=0$ or $(x-4)=0$
i.e $x = -1$ and $x = 4$
Therefore, –1 and 4 are zeroes of the quadratic polynomial. Also note from Fig. 2.2 that –1 and 4 are the x-coordinates of the points where the graph of $y = x^2– 3x – 4$ intersects the x-axis. Thus, the zeroes of the quadratic polynomial $x^2 – 3x – 4$ are x-coordinates of the points where the graph of $y = x^2 – 3x – 4 $ intersects the x-axis.
This fact is true for any quadratic polynomial, i.e., the zeroes of a quadratic polynomial $ax^2 + bx + c$, $a ≠ 0$, are precisely the x-coordinates of the points where the parabola representing $y = ax^2 + bx + c$ intersects the x-axis.