Question

Draw the graphs of the quadratic polynomial
$f\left(x\right)=3-2x-x^2$

Answer 1

Let y =f(x) or, $y=3-2x-x^2$.

Let us list a few values of $y=3-2x-x^2$

corresponding to a few values of x as follows

$x$	$-5$	$-4$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3/4$
$y=3-2x-x^2$	$-12$	$-5$	$0$	$3$	$4$	$3$	$0$	$-5$	$-12/-21$

Thus, the. following points lie on the graph of the polynomial $y=3-2x-x^2$

$(-5,-12),(-4,-5),(-3,0),(-2,3),(-1,4),(0,3),(1,0),(2,-5),(3,-12)$ 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 $y=3-2x-x^2$. The curve thus obtained represents a parabola, as shown in figure. The highest point P $(-1,3)$, 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.