1. The problem involves analyzing and graphing four quadratic functions: $f(x) = -x^2 - 1$, $g(x) = -2x^2 - 18x - 36$, $h(x) = x^2 + 5$, and $k(x) = x^2 + 14x + 48$. 2. Recall the standard form of a quadratic function is $ax^2 + bx + c$. 3. The vertex form is $a(x-h)^2 + k$, where $(h,k)$ is the vertex. 4. For $g(x)$ and $k(x)$, factorization is given: $$g(x) = -2(x+6)(x+3)$$ $$k(x) = (x+8)(x+6)$$ 5. The vertex of a parabola $ax^2 + bx + c$ is at $x = -\frac{b}{2a}$. 6. Calculate vertices: - For $f(x) = -x^2 - 1$, $a = -1$, $b=0$, vertex $x = -\frac{0}{2(-1)}=0$, vertex point $(0, f(0)) = (0, -1)$. - For $g(x) = -2x^2 - 18x - 36$, $a = -2$, $b = -18$, vertex $x = -\frac{-18}{2(-2)} = -\frac{-18}{-4} = -4.5$, vertex $y = g(-4.5) = -2(-4.5)^2 - 18(-4.5) - 36 = -2(20.25) + 81 - 36 = -40.5 + 81 - 36 = 4.5$. - For $h(x) = x^2 + 5$, $a=1$, $b=0$, vertex $x=0$, vertex point $(0,5)$. - For $k(x) = x^2 + 14x + 48$, $a=1$, $b=14$, vertex $x = -\frac{14}{2} = -7$, vertex $y = k(-7) = (-7)^2 + 14(-7) + 48 = 49 - 98 + 48 = -1$. 7. The x-intercepts are roots of the quadratic equations: - $f(x)$ has no real roots since $-x^2 - 1 = 0$ implies $x^2 = -1$ (no real solution). - $g(x)$ roots from factorization: $x = -6, -3$. - $h(x)$ has no real roots since $x^2 + 5 = 0$ implies $x^2 = -5$ (no real solution). - $k(x)$ roots from factorization: $x = -8, -6$. 8. Summary: - $f(x)$: downward parabola, vertex at $(0,-1)$, no x-intercepts. - $g(x)$: downward parabola, vertex at $(-4.5,4.5)$, x-intercepts at $-6$ and $-3$. - $h(x)$: upward parabola, vertex at $(0,5)$, no x-intercepts. - $k(x)$: upward parabola, vertex at $(-7,-1)$, x-intercepts at $-8$ and $-6$. Final answer: The functions are correctly analyzed with vertices and intercepts as above.