1. Problem: Analyze $f(x) = x^{2} + 10x + 24$. 2. Find the vertex: Use vertex formula $x = -\frac{b}{2a}$ for $a=1$, $b=10$: $$x = -\frac{10}{2 \times 1} = -5$$ Then find $y$ at $x=-5$: $$f(-5) = (-5)^2 + 10(-5) + 24 = 25 - 50 + 24 = -1$$ Vertex is $(-5, -1)$. 3. Find the roots: Solve $x^{2} + 10x + 24 = 0$ by factoring: $$x^{2} + 6x + 4x + 24 = 0 \Rightarrow (x+6)(x+4) = 0$$ So roots are $x = -6$ and $x = -4$. 4. Axis of symmetry: The vertical line through vertex $x = -5$. 5. Domain: All real numbers $(-\infty, \infty)$ because it is a quadratic. 6. Range: Since parabola opens upwards ($a=1>0$), range is $[-1, \infty)$. --- 1. Problem: Analyze $f(x) = x^{2} - 6x + 8$. 2. Find the vertex: $$x = -\frac{-6}{2 \times 1} = 3$$ Calculate $y$: $$f(3) = 3^{2} - 6 \times 3 + 8 = 9 - 18 + 8 = -1$$ Vertex is $(3, -1)$. 3. Find the roots: Factor: $$x^{2} - 6x + 8 = (x - 2)(x - 4) = 0$$ Roots are $x = 2$ and $x = 4$. 4. Axis of symmetry: $x = 3$. 5. Domain: $(-\infty, \infty)$ 6. Range: Since $a=1>0$, range is $[-1, \infty)$. --- Graph shapes: Two parabolas both opening upwards. - First parabola vertex at $(-5, -1)$ with roots $-6$ and $-4$. - Second parabola vertex at $(3, -1)$ with roots $2$ and $4$. They share the same minimum $y$-value $-1$ but occur at different $x$ values. ---