(i) `y=x^(2)+x+1`
Here the quadratic expression does not have real factors.
So using completing square method, we have
`" "y= (x+ (1)/(2))^(2)+ (3)/(4)`
or `" "y- (3)/(4)= (x+ (1)/(2))^(2)`
Thus, the vertex of the parabola is `(-(1)/(2), (3)/(4))` and the graph is concave upwards as shown in the adjacent figure.
(ii) `y= x^(2)-2x-3`
We have `y=x^(2)-2x-3= (x-3)(x+1)`
Thus, the graph of the function is a parabola opening upwards, intersecting the `x`-axis at `x=-1 and x=3`. Also vertex is `(1, -4)`. Therefore, the graph of the function is as shown in the following figure.
(iii) `y= 2+x-x^(2)`
We have `y=2+x-x^(2)=-(x-2)(x+1)`
Thus, the graph of the function is a parabola opening downwards, intersecting the `x`-axis at `x=-1 and x =2`.
Also the vertex is `((1)/(2), (9)/(4))`.
Therefore, the graph of the function is as shown in the following figure.
(iv) `y=x-1-x^(2)`
Here the quadratic expression does not have real factors.
So using the completing square method, we have `-y=x^(2)-x+1`
or `" "-y= (x- (1)/(2))^(2)+ (3)/(4) or - (y+ (3)/(4))= (x+ (1)/(2))^(2)`
Thus, the vertex of the parabla is `(-(1)/(2), - (3)/(4)`, and the graph is concave downwards. Therefore, the graph of the function is as shown in the following figure.
