Quadratic Parabola 619896

1. **State the problem:** We are given the quadratic function $$y = x^2 - 1$$ and need to complete a table of values, plot the points, and identify key features of the parabola. 2. **Complete the table of values:** Use the formula $$y = x^2 - 1$$ to find $y$ for each $x$. - For $x = -2$: $$y = (-2)^2 - 1 = 4 - 1 = 3$$ - For $x = -1$: $$y = (-1)^2 - 1 = 1 - 1 = 0$$ - For $x = 0$: $$y = 0^2 - 1 = 0 - 1 = -1$$ - For $x = 1$: $$y = 1^2 - 1 = 1 - 1 = 0$$ - For $x = 2$: $$y = 2^2 - 1 = 4 - 1 = 3$$ Completed table: | $x$ | -2 | -1 | 0 | 1 | 2 | |-----|----|----|---|---|---| | $y$ | 3 | 0 | -1| 0 | 3 | 3. **Plot the points:** Plot the points $(-2,3)$, $(-1,0)$, $(0,-1)$, $(1,0)$, and $(2,3)$ on a coordinate plane. 4. **State the features:** i. **Type of turning point:** Since the coefficient of $x^2$ is positive, the parabola opens upwards, so the turning point is a minimum. ii. **Axis of symmetry:** The parabola is symmetric about the vertical line through the vertex, which is $x=0$. iii. **Coordinates of the turning point:** The vertex is at $(0, -1)$. iv. **Coordinates of the y-intercept:** When $x=0$, $y=-1$, so the y-intercept is $(0, -1)$. v. **Coordinates of the x-intercepts:** Solve $x^2 - 1 = 0$: $$x^2 = 1$$ $$x = \pm 1$$ So the x-intercepts are $(-1, 0)$ and $(1, 0)$. **Final answers:** - Table: $y$ values are 3, 0, -1, 0, 3 for $x = -2, -1, 0, 1, 2$ respectively. - Turning point: minimum at $(0, -1)$. - Axis of symmetry: $x=0$. - y-intercept: $(0, -1)$. - x-intercepts: $(-1, 0)$ and $(1, 0)$. This completes the tasks.