1. **State the problem:** We are given a parabola with vertex at $(-1,-4)$, crossing the y-axis at $(0,3)$, and x-axis at approximately $(-3,0)$ and $(1,0)$. We need to find: (a) $h(2)$, the value of the function at $x=2$. (b) One value of $x$ such that $h(x) = -2$. 2. **Find the function formula:** Since the parabola opens upwards and has vertex $(-1,-4)$, its equation can be written in vertex form: $$h(x) = a(x+1)^2 - 4$$ We use the point $(0,3)$ to find $a$: $$3 = a(0+1)^2 - 4$$ $$3 = a(1)^2 - 4$$ $$3 = a - 4$$ $$a = 7$$ So the function is: $$h(x) = 7(x+1)^2 - 4$$ 3. **Find $h(2)$:** $$h(2) = 7(2+1)^2 - 4 = 7(3)^2 - 4 = 7 \times 9 - 4 = 63 - 4 = 59$$ 4. **Find $x$ such that $h(x) = -2$:** Set the function equal to $-2$: $$7(x+1)^2 - 4 = -2$$ Add 4 to both sides: $$7(x+1)^2 = -2 + 4$$ $$7(x+1)^2 = 2$$ Divide both sides by 7: $$\frac{7(x+1)^2}{\cancel{7}} = \frac{2}{\cancel{7}}$$ $$ (x+1)^2 = \frac{2}{7}$$ Take the square root of both sides: $$x+1 = \pm \sqrt{\frac{2}{7}}$$ Solve for $x$: $$x = -1 \pm \sqrt{\frac{2}{7}}$$ One value of $x$ is: $$x = -1 + \sqrt{\frac{2}{7}}$$ **Final answers:** (a) $h(2) = 59$ (b) One value of $x$ for which $h(x) = -2$ is $x = -1 + \sqrt{\frac{2}{7}}$