Question

graph the following function. estimate the intervals on which the function is increasing or decreasing and any relative maxima or minima.\\( f(x) = 8x^2 \\)\\( \dots \\)\\( \text{use the graphing tool to graph the function.} \\)\\( \text{click to enlarge graph} \\)

Explanation:

Step1: Analyze the function type

The function \( f(x) = 8x^2 \) is a quadratic function in the form \( f(x)=ax^2+bx + c \) (here \( b = 0 \), \( c = 0 \), \( a=8>0 \)). The graph of a quadratic function \( y = ax^2+bx + c \) with \( a>0 \) is a parabola opening upwards. The vertex of the parabola \( y=ax^2+bx + c \) is at \( x=-\frac{b}{2a} \). For \( f(x)=8x^2 \), \( b = 0 \), so the vertex is at \( x = 0 \) (the y - axis), and the vertex point is \( (0,0) \).

Step2: Determine increasing and decreasing intervals

To find where the function is increasing or decreasing, we can use the concept of the derivative (or the properties of the parabola). The derivative of \( f(x)=8x^2 \) is \( f^\prime(x)=16x \).

• When \( f^\prime(x)<0 \), the function is decreasing. Solve \( 16x<0 \), we get \( x < 0 \). So the function is decreasing on the interval \( (-\infty,0) \).
• When \( f^\prime(x)>0 \), the function is increasing. Solve \( 16x>0 \), we get \( x>0 \). So the function is increasing on the interval \( (0,\infty) \).

Step3: Determine relative maxima and minima

Since the parabola opens upwards (because \( a = 8>0 \)), the vertex is the minimum point. There is no relative maximum because the parabola extends upwards to infinity. The relative minimum occurs at \( x = 0 \), and \( f(0)=8\times0^2 = 0 \), so the relative minimum is \( 0 \) at \( x = 0 \), and there is no relative maximum.

Answer:

• The function \( f(x)=8x^2 \) is decreasing on the interval \( (-\infty,0) \) and increasing on the interval \( (0,\infty) \).
• The function has a relative minimum of \( 0 \) at \( x = 0 \) and no relative maximum. The graph is a parabola opening upwards with vertex at \( (0,0) \).