1. **State the problem:** We need to find the range of each quadratic function given either by formula or table of values. 2. **Recall the general form and range of a quadratic function:** A quadratic function is generally written as $$y = a(x-h)^2 + k$$ where $(h,k)$ is the vertex. - If $a > 0$, the parabola opens upward and the range is $$[k, \infty)$$. - If $a < 0$, the parabola opens downward and the range is $$(-\infty, k]$$. 3. **Analyze each function:** - For $y = -x^2$ (H and R): - Vertex at $(0,0)$, $a = -1 < 0$ so parabola opens downward. - Range is $$(-\infty, 0]$$. - For $y = -x^2 + 3$ (U): - Vertex at $(0,3)$, $a = -1 < 0$. - Range is $$(-\infty, 3]$$. - For $y = x^2 - 3$ (S): - Vertex at $(0,-3)$, $a = 1 > 0$. - Range is $$[-3, \infty)$$. - For $y = (x + 4)^2 + 3$ (E): - Vertex at $(-4,3)$, $a = 1 > 0$. - Range is $$[3, \infty)$$. - For $y = -2(x - 1)^2 - 3$ (T bottom-left): - Vertex at $(1,-3)$, $a = -2 < 0$. - Range is $$(-\infty, -3]$$. - For $y = (x - 3)^2$ (H bottom-center): - Vertex at $(3,0)$, $a = 1 > 0$. - Range is $$[0, \infty)$$. 4. **For tables, find the minimum or maximum y-value:** - For A (table): x = 1,2,3,4,5; y = 11,6,3,2,3 - Minimum y is 2, maximum y is 11. - Since y decreases then increases, vertex at x=4 with y=2. - Range is $$[2, 11]$$. - For T (bottom-right table): x = 0,2,4,6,8; y = 8,4,8,20,40 - Minimum y is 4 at x=2. - Range is $$[4, \infty)$$ since values increase after minimum. 5. **Summary of ranges:** - H, R: $$(-\infty, 0]$$ - U: $$(-\infty, 3]$$ - A: $$[2, 11]$$ - S: $$[-3, \infty)$$ - E: $$[3, \infty)$$ - T (bottom-left): $$(-\infty, -3]$$ - H (bottom-center): $$[0, \infty)$$ - T (bottom-right): $$[4, \infty)$$ This completes finding the ranges for all given quadratic functions.