Question

graph the function f(x) = -2x². plot the vertex. then plot another point on the parabola. if you make a mistake, you can erase your parabola by selecting the second point and placing it on top of the first.

Explanation:

Step1: Identify the vertex of the parabola

For a quadratic function in the form \( f(x) = ax^2 + bx + c \), the vertex is at \( (-\frac{b}{2a}, f(-\frac{b}{2a})) \). For \( f(x) = -2x^2 \), \( a = -2 \), \( b = 0 \), \( c = 0 \). So the x - coordinate of the vertex is \( -\frac{0}{2\times(-2)} = 0 \). Substitute \( x = 0 \) into the function: \( f(0)=-2\times0^2 = 0 \). So the vertex is \( (0, 0) \).

Step2: Find another point on the parabola

Let's choose \( x = 1 \). Substitute \( x = 1 \) into \( f(x)=-2x^2 \), we get \( f(1)=-2\times1^2=-2 \). So the point \( (1, -2) \) is on the parabola. Since the parabola is symmetric about the y - axis, the point \( (-1, -2) \) is also on the parabola. The parabola opens downward because \( a=-2<0 \). To graph the parabola, plot the vertex \( (0, 0) \) and then plot the point \( (1, -2) \) (or \( (-1, -2) \)) and draw a parabola opening downward passing through these points.

Answer:

The vertex is at \( (0, 0) \), and another point (e.g., \( (1, -2) \)) can be plotted to draw the parabola \( f(x)=-2x^2 \) which opens downward.