8. sketch a parabola that has x - intercepts of - 6 and 2 and a range of y ≥ - 8.\n9. consider the function…

Explanation:

Step1: Determine the type of function

For a quadratic function (y = ax^{2}+bx + c), if (a>0), the parabola opens upwards and has a minimum value; if (a < 0), it opens downwards and has a maximum value. For (f(x)=x^{2}-2x - 3), where (a = 1>0), so the function has a minimum value.

Step2: Find the vertex (to get the minimum value)

The (x) - coordinate of the vertex of a quadratic function (y=ax^{2}+bx + c) is given by (x=-\frac{b}{2a}). Here, (a = 1), (b=-2). Then (x=-\frac{-2}{2\times1}=1). Substitute (x = 1) into the function (f(x)): (f(1)=1^{2}-2\times1 - 3=1 - 2-3=-4)

Answer:

a. The function (f(x)=x^{2}-2x - 3) has a minimum value, and the minimum value is (-4).

b. To graph the function (y=x^{2}-2x - 3=(x - 1)^{2}-4):

• The vertex is ((1,-4)).
• The (y) - intercept: set (x = 0), then (y=-3).
• The (x) - intercepts: set (y = 0), (x^{2}-2x - 3=0), factoring gives ((x - 3)(x + 1)=0), so (x=3) or (x=-1). Plot the vertex ((1,-4)), (y) - intercept ((0,-3)), (x) - intercepts ((3,0)) and ((-1,0)) and draw a parabola opening upwards.