Quadratic Table A003B6

1. **State the problem:** We are given the quadratic function $$y = x^2 + 2x - 3$$ and need to complete a table of values, plot the parabola, and identify key features. 2. **Complete the table of values:** Calculate $y$ for each $x$: - For $x = -4$: $$y = (-4)^2 + 2(-4) - 3 = 16 - 8 - 3 = 5$$ - For $x = -3$: $$y = (-3)^2 + 2(-3) - 3 = 9 - 6 - 3 = 0$$ - For $x = -2$: $$y = (-2)^2 + 2(-2) - 3 = 4 - 4 - 3 = -3$$ - For $x = -1$: $$y = (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4$$ - For $x = 0$: $$y = 0^2 + 2(0) - 3 = 0 + 0 - 3 = -3$$ - For $x = 1$: $$y = 1^2 + 2(1) - 3 = 1 + 2 - 3 = 0$$ - For $x = 2$: $$y = 2^2 + 2(2) - 3 = 4 + 4 - 3 = 5$$ Table: | x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | |---|----|----|----|----|---|---|---| | y | 5 | 0 | -3 | -4 | -3| 0 | 5 | 3. **Plot the points:** Draw axes with $x$ from -5 to 3 and $y$ from -5 to 6. Plot points $(-4,5)$, $(-3,0)$, $(-2,-3)$, $(-1,-4)$, $(0,-3)$, $(1,0)$, $(2,5)$ to form a parabola opening upwards. 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 axis of symmetry is given by $$x = -\frac{b}{2a} = -\frac{2}{2 \times 1} = -1$$ - iii. Coordinates of the turning point: Substitute $x = -1$ into the equation: $$y = (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4$$ So the turning point is $(-1, -4)$. - iv. Coordinates of the y-intercept: Set $x=0$: $$y = 0 + 0 - 3 = -3$$ So the y-intercept is $(0, -3)$. - v. Coordinates of the x-intercepts: Solve $x^2 + 2x - 3 = 0$. Factor: $$(x + 3)(x - 1) = 0$$ So $x = -3$ or $x = 1$. The x-intercepts are $(-3, 0)$ and $(1, 0)$. **Final answer:** - Table completed. - Parabola plotted with points. - Turning point: minimum at $(-1, -4)$. - Axis of symmetry: $x = -1$. - Y-intercept: $(0, -3)$. - X-intercepts: $(-3, 0)$ and $(1, 0)$.