Question

A square has a side length of X centimeters. One dimension increases by 4 cm and the other dimension increases by 2 cm, forming a new rectangle.
a. Make a sketch of the new rectangle.

b. Write two expressions for the area of the new rectangle, one in factored form and one in expanded form.

c. Write one equation for the Area, A, of the rectangle.

d. Graph the equation on the coordinate grid below by finding: x-intercepts, y-intercept, line of symmetry, and the vertex (ordered pair). Show your work.

e. Does the graph open upward or downward?

How do you know from the equation?​

Answer 1

(a) Look at the picture below: we started with the square ABCD, with side length $x$, and we built the rectangle AEFG such that

$AE=AB+4=x+4,\quad AG=AC+2=x+2$

(b) Since the dimensions of AEFG are $x+4$ and $x+2$, its area will be the product of its dimensions:

$A=(x+2)(x+4)=x^2+6x+8$

(c) The area $A(x)$ is a function of the side length $x$:

$A(x)=x^2+6x+8$

(d) Since $A(x)$ is a polynomial with degree 2, it represents a parabola. In order to find the x intercept, we must set $y=0$:

$x^2+6x+8=0 \iff x=-2\lor x=-4$

Similarly, in order to find the y intercept, we must set $x=0$:

$A(0)=0^2+6\cdot 0+8=8$

The x-coordinate of the vertex is given by

$x=-\dfrac{b}{2a}=-\dfrac{6}{2}=-3$

And its y coordinate is given by

$A(-3)=(-3)^2+6\cdot (-3)+8=9-18+8=-1$

The line of symmetry is the vertical line passing through the vertex, so it must be

$x=-3$

(see the second picture for the graph)

(e) The graph opens upward. We knew that from the equation, because given a parabola

$y=ax^2+bx+c=0$

its graph will open upwards if $a>0$, downwards if $a$. In your case, $a=1$, so it's positive and the parabola opens upwards.