How to Sketch the Graph of yx^2^2-1 in the Domain [-3, 0]

The equation yx^2^2-1 represents a parabola. To sketch its graph in the domain [-3, 0], we will follow several steps involving finding the vertex, x-intercepts, and understanding the behavior of the function within the given domain.

Understanding the Equation

The given equation is yx^2^2-1. This equation is in the form of a parabola opened upward, as the coefficient of x^2 is 1, which is positive. The vertex form of a parabola is ya(x-h)^2 k where the vertex is at (h, k).

Identifying the Vertex

The vertex of the parabola yx^2^2-1 can be determined by rewriting the equation in the vertex form:

y (x - 0)^2 - 1, which simplifies to y x^2 - 1. Therefore, the vertex is at (0, -1).

Domain and x-Intercepts

We are only interested in sketching the graph in the domain [-3, 0]. To find the x-intercepts (where the graph crosses the x-axis), we set y 0 and solve for x:

0 x^2 - 1

x^2 1

x ±1

Only the intercept within the domain will be considered, which is x -1. At x -1, we get

y (-1)^2 - 1 1 - 1 0.

Plotting the Key Points

Let's find the y-values for x-values -3, 0, and -1:

x -3:

y (-3)^2 - 1 9 - 1 8

x 0:

y (0)^2 - 1 -1

x -1:

y (-1)^2 - 1 0

So, we have the points: (-3, 8), (-1, 0), and (0, -1).

Sketching the Curve

To sketch the curve, we will connect the points with a smooth curve that opens upward from the vertex. The curve should pass through the points (-3, 8), (-1, 0), and (0, -1). The curve will have a minimum point at (-1, 0).

Summary of Key Points

Endpoints of the Domain: x -3, y 8 x 0, y -1

x-Intercept: x -1, y 0

Minimum Point: x -1, y 0

Vertex: (0, -1)

Conclusion

By connecting these points with a smooth curve, we can accurately sketch the graph of yx^2^2-1 within the domain [-3, 0]. The curve will start at (-3, 8), pass through (-1, 0), and end at (0, -1).