Question

1. for $y = x^2 - 2x + 1$, what is the vertex and does the parabola open upwards?

a. vertex: $(1,0)$, no
b. vertex: $(0,1)$, yes
c. vertex: $(1,0)$, yes
d. vertex: $(0,1)$, no

1. what is the domain of any quadratic function?

a. $1,\infty)$
b. $0,\infty)$
c. $0,1$
d. $(-\infty,\infty)$

Explanation:

Step1: Find vertex x-coordinate

For $y=ax^2+bx+c$, $x=-\frac{b}{2a}$. Here $a=1, b=-2$, so:
$x=-\frac{-2}{2(1)}=1$

Step2: Find vertex y-coordinate

Substitute $x=1$ into $y=x^2-2x+1$:
$y=(1)^2-2(1)+1=1-2+1=0$

Step3: Check parabola direction

Since $a=1>0$, parabola opens upwards.

Step4: Identify quadratic domain

Quadratic functions accept all real $x$, so domain is $(-\infty, \infty)$.

Answer:

1. c. Vertex: (1,0), Yes
2. d. $(-\infty, \infty)$