Question

graph the parabola represented by the following equation:
$f(x) = 2x^2 - 4x - 6$
plot the vertex first, then plot a second point on the parabola.

Explanation:

Step1: Find the vertex of the parabola

The general form of a quadratic function is \( f(x) = ax^2 + bx + c \), and the x - coordinate of the vertex is given by \( x=-\frac{b}{2a} \). For the function \( f(x)=2x^{2}-4x - 6 \), we have \( a = 2 \), \( b=-4 \), and \( c=-6 \).
Using the formula for the x - coordinate of the vertex: \( x=-\frac{-4}{2\times2}=\frac{4}{4} = 1 \).
To find the y - coordinate of the vertex, we substitute \( x = 1 \) into the function: \( f(1)=2\times(1)^{2}-4\times(1)-6=2 - 4-6=-8 \). So the vertex of the parabola is at the point \( (1,-8) \).

Step2: Find a second point on the parabola

Let's choose \( x = 0 \) (we can choose any value of x, x = 0 is a simple choice). Substitute \( x = 0 \) into the function \( f(x)=2x^{2}-4x - 6 \):
\( f(0)=2\times(0)^{2}-4\times(0)-6=-6 \). So the point \( (0,-6) \) lies on the parabola.

To graph the parabola, we first plot the vertex \( (1,-8) \) and then plot the point \( (0,-6) \) (or any other point we find).

Answer:

The vertex of the parabola \( f(x)=2x^{2}-4x - 6 \) is \( (1,-8) \), and a second point on the parabola (for example) is \( (0,-6) \). When graphing, plot \( (1,-8) \) first and then \( (0,-6) \) (or another valid point) on the coordinate plane.