A curve is given by \(y = 5 - x - 2x^{2}\). Find the equation of its line of symmetry.
Answer Details
The line of symmetry of a parabola is a vertical line that passes through the vertex of the parabola. The equation of the line of symmetry for a parabola of the form \(y = a(x - h)^2 + k\) is simply the vertical line \(x = h\).
The given curve is a quadratic function of the form \(y = -2x^2 - x + 5\), which is a downward-facing parabola. The line of symmetry for this parabola is the vertical line that passes through its vertex.
To find the vertex of the parabola, we can first find the x-coordinate of the vertex using the formula \(x = \frac{-b}{2a}\). In this case, \(a = -2\) and \(b = -1\), so we have:
$$x = \frac{-b}{2a} = \frac{-(-1)}{2(-2)} = \frac{1}{4}$$
To find the corresponding y-coordinate, we can substitute this value of x into the equation for the curve:
$$y = -2x^2 - x + 5 = -2\left(\frac{1}{4}\right)^2 - \frac{1}{4} + 5 = \frac{41}{8}$$
Therefore, the vertex of the parabola is \(\left(\frac{1}{4}, \frac{41}{8}\right)\), and the equation of its line of symmetry is simply:
$$x = \frac{1}{4}$$
So the correct option is \(\boxed{\text{(c) } x = \frac{1}{4}}\).