Question

1. sketch a parabola that has x - intercepts of - 6 and 2 and a range of y ≥ - 8.
2. consider the function f(x)=x² - 2x - 3.

a. determine whether the function has a maximum or minimum value. then find the value.
b. graph 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.