Question

question 9 (3 points)
graph the function:
$f(x) = 2x^2 + 1$
on which interval is the function increasing?
a $(-\infty, 0)$
b $(0, \infty)$
c $(-\infty, \infty)$
d $(-1, 1)$

Explanation:

Step1: Identify the function type

The function \( f(x) = 2x^2 + 1 \) is a quadratic function in the form \( f(x)=ax^2 + bx + c \), where \( a = 2 \), \( b = 0 \), \( c = 1 \). Since \( a=2>0 \), the parabola opens upward.

Step2: Find the vertex

For a quadratic function \( f(x)=ax^2+bx + c \), the x - coordinate of the vertex is given by \( x=-\frac{b}{2a} \). Here, \( b = 0 \) and \( a=2 \), so \( x =-\frac{0}{2\times2}=0 \). The vertex of the parabola is at \( (0,f(0)) \). Substituting \( x = 0 \) into the function, we get \( f(0)=2(0)^2+1 = 1 \), so the vertex is \( (0,1) \).

Step3: Determine the increasing interval

For a parabola that opens upward ( \( a>0 \) ), the function is decreasing on the interval \( (-\infty, h) \) and increasing on the interval \( (h,\infty) \), where \( h \) is the x - coordinate of the vertex. Here, \( h = 0 \), so the function is increasing on the interval \( (0,\infty) \).

Answer:

b. \( (0, \infty) \)