(a) Look at the picture below: we started with the square ABCD, with side length [tex]x[/tex], and we built the rectangle AEFG such that
[tex]AE=AB+4=x+4,\quad AG=AC+2=x+2[/tex]
(b) Since the dimensions of AEFG are [tex]x+4[/tex] and [tex]x+2[/tex], its area will be the product of its dimensions:
[tex]A=(x+2)(x+4)=x^2+6x+8[/tex]
(c) The area [tex]A(x)[/tex] is a function of the side length [tex]x[/tex]:
[tex]A(x)=x^2+6x+8[/tex]
(d) Since [tex]A(x)[/tex] is a polynomial with degree 2, it represents a parabola. In order to find the x intercept, we must set [tex]y=0[/tex]:
[tex]x^2+6x+8=0 \iff x=-2\lor x=-4[/tex]
Similarly, in order to find the y intercept, we must set [tex]x=0[/tex]:
[tex]A(0)=0^2+6\cdot 0+8=8[/tex]
The x-coordinate of the vertex is given by
[tex]x=-\dfrac{b}{2a}=-\dfrac{6}{2}=-3[/tex]
And its y coordinate is given by
[tex]A(-3)=(-3)^2+6\cdot (-3)+8=9-18+8=-1[/tex]
The line of symmetry is the vertical line passing through the vertex, so it must be
[tex]x=-3[/tex]
(see the second picture for the graph)
(e) The graph opens upward. We knew that from the equation, because given a parabola
[tex]y=ax^2+bx+c=0[/tex]
its graph will open upwards if [tex]a>0[/tex], downwards if [tex]a<0[/tex]. In your case, [tex]a=1[/tex], so it's positive and the parabola opens upwards.
