Question

function a gives the area, in square inches, of a square with side length x inches.
a. complete the table.

x	0	1	2	3	4	5	6
a(x)

b. represent function a using an equation.
c. sketch a graph of function a.

Explanation:

Part a

Step1: Recall the area formula for a square

The area \( A \) of a square with side length \( x \) is given by \( A = x^2 \).

Step2: Calculate \( A(x) \) for each \( x \)

• For \( x = 0 \): \( A(0)=0^2 = 0 \)
• For \( x = 1 \): \( A(1)=1^2 = 1 \)
• For \( x = 2 \): \( A(2)=2^2 = 4 \)
• For \( x = 3 \): \( A(3)=3^2 = 9 \)
• For \( x = 4 \): \( A(4)=4^2 = 16 \)
• For \( x = 5 \): \( A(5)=5^2 = 25 \)
• For \( x = 6 \): \( A(6)=6^2 = 36 \)

Part b

Step1: Recall the area formula for a square

The area of a square is calculated by squaring the length of its side. If the side length is \( x \) (in inches) and the area is \( A(x) \) (in square inches), the formula that relates them is the area formula for a square, which is \( A = \text{side}^2 \).

Step2: Write the function

Substituting the side length \( x \) into the formula, we get the function \( A(x)=x^2 \).

Part c

Step1: Identify the type of function

The function \( A(x)=x^2 \) is a quadratic function, and the graph of a quadratic function \( y = ax^2+bx + c \) (in this case \( a = 1 \), \( b = 0 \), \( c = 0 \)) is a parabola. Since \( a>0 \), the parabola opens upwards.

Step2: Determine the vertex

For the quadratic function \( y=x^2 \), the vertex is at \((h,k)\) where \( h=-\frac{b}{2a}\) and \( k = A(h) \). Here, \( b = 0 \), \( a = 1 \), so \( h = 0 \) and \( k=0^2 = 0 \). So the vertex is at \((0,0)\).

Step3: Plot key points

Using the values from part a, we have points \((0,0)\), \((1,1)\), \((2,4)\), \((3,9)\), \((4,16)\), \((5,25)\), \((6,36)\). Plot these points on the coordinate plane with \( x \)-coordinate as the side length and \( y \)-coordinate as the area.

Step4: Draw the graph

Draw a smooth curve through these plotted points to represent the parabola \( y = x^2 \).

Answer:

\( x \)	0	1	2	3	4	5	6