Question

the graph of ( f(x)=x^{2} ) has which of the following key features?
a. a constant first difference
b. a horizontal asymptote
c. a turning point at ( (1,1) )
d. a vertex at ( (0,0) )

Explanation:

Step1: Analyze the function \(y = x^{2}\)

The function \(y=x^{2}\) is a quadratic function. The general form of a quadratic function is \(y = ax^{2}+bx + c\). For \(y=x^{2}\), \(a = 1\), \(b=0\), \(c = 0\).

Step2: Find the vertex

The \(x\) - coordinate of the vertex of a quadratic function \(y=ax^{2}+bx + c\) is given by \(x=-\frac{b}{2a}\). Substituting \(a = 1\) and \(b = 0\), we get \(x = 0\). Substituting \(x = 0\) into \(y=x^{2}\), we get \(y=0\). So the vertex is \((0,0)\).

Step3: Check for horizontal asymptote

A horizontal asymptote exists for rational functions (e.g., \(y=\frac{f(x)}{g(x)}\) where \(\text{deg}(f(x))\leq\text{deg}(g(x))\)). Since \(y=x^{2}\) is a polynomial function (not a rational function in the form that would have a horizontal asymptote), there is no horizontal asymptote.

Step4: Check for turning point

The first - derivative of \(y=x^{2}\) is \(y^{\prime}=2x\). Setting \(y^{\prime}=0\), we get \(2x = 0\Rightarrow x = 0\). The second - derivative \(y^{\prime\prime}=2>0\). So the function has a minimum (a turning point) at \(x = 0\), \(y=0\) (not at \((1,1)\)).

Step5: Check for constant first - difference

For a linear function \(y=mx + b\), the first - difference \(\Delta y=m\) (constant). For \(y=x^{2}\), if \(x_1\) and \(x_2\) are two values of \(x\), \(y_1=x_1^{2}\) and \(y_2=x_2^{2}\), \(\Delta y=x_2^{2}-x_1^{2}=(x_2 - x_1)(x_2 + x_1)\), which is not constant.

Answer:

d. A vertex at \((0,0)\)