Question

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

Explanation:

Step1: Analyze option a

The function $f(x)=x^2$ is a parabola. Its turning point (vertex) occurs where the derivative is 0. The derivative $f'(x)=2x$. Setting $2x=0$ gives $x=0$, so $f(0)=0^2=0$. The turning point is $(0,0)$, not $(1,1)$, so a is incorrect.

Step2: Analyze option b

First differences for a quadratic function are linear, not constant. For $f(x)=x^2$, take $x=0,1,2,3$: $f(0)=0$, $f(1)=1$, $f(2)=4$, $f(3)=9$. First differences are $1-0=1$, $4-1=3$, $9-4=5$, which are not constant, so b is incorrect.

Step3: Analyze option c

A horizontal asymptote exists if $\lim_{x\to\pm\infty}f(x)$ is a finite number. $\lim_{x\to\pm\infty}x^2=\infty$, so there is no horizontal asymptote, c is incorrect.

Step4: Analyze option d

The standard form of a parabola is $f(x)=a(x-h)^2+k$, where $(h,k)$ is the vertex. For $f(x)=x^2=(x-0)^2+0$, $h=0$, $k=0$, so the vertex is $(0,0)$, d is correct.

Answer:

d. A vertex at (0, 0)