The problem asks us to sketch the graphs of the following two quadratic functions: (1) $y = x^2 + 1$ (2) $y = -2x^2 - 1$

1. Problem Description

The problem asks us to sketch the graphs of the following two quadratic functions:

(1) $y = x^2 + 1$

(2) $y = -2x^2 - 1$

2. Solution Steps

(1) For the function $y = x^2 + 1$, this is a parabola.

The basic parabola $y = x^2$ has its vertex at the origin $(0, 0)$.

The graph of $y = x^2 + 1$ is the graph of $y = x^2$ shifted upwards by 1 unit. Therefore, the vertex of this parabola is $(0, 1)$.

The parabola opens upwards since the coefficient of $x^2$ is positive (1).

(2) For the function $y = -2x^2 - 1$, this is also a parabola.

The basic parabola $y = x^2$ has its vertex at the origin $(0, 0)$.

The graph of $y = -2x^2$ is the graph of $y = x^2$ reflected over the x-axis and vertically stretched by a factor of

2. So, $y = -2x^2$ opens downwards and is narrower than $y=x^2$.

The graph of $y = -2x^2 - 1$ is the graph of $y = -2x^2$ shifted downwards by 1 unit. Therefore, the vertex of this parabola is $(0, -1)$.

The parabola opens downwards since the coefficient of $x^2$ is negative (-2).

3. Final Answer

The graphs are parabolas.

(1) $y = x^2 + 1$: Parabola with vertex at $(0, 1)$, opening upwards.

(2) $y = -2x^2 - 1$: Parabola with vertex at $(0, -1)$, opening downwards and narrower than $y = x^2$.

Since the problem requires us to sketch the graphs, instead of just plotting a graph, it would involve plotting some specific points and indicating the parabola shape with the appropriate vertex and direction of opening. A detailed, scaled plot would require more tools than simple text can provide.