1. **Stating the problem:** We have a quadratic function $y=f(x)$ with a tangent line $y=-1$ to the curve. The vertex is at $(0,1)$ and the curve passes through $(2,-1)$ and $(-3,14)$. 2. **Axis of symmetry:** The axis of symmetry of a parabola $y=f(x)$ with vertex $(h,k)$ is the vertical line $x=h$. Since the vertex is at $(0,1)$, the axis of symmetry is: $$x=0$$ 3. **Expressing $f(x)$ in vertex form:** The vertex form of a quadratic is: $$f(x)=a(x+b)^2+c$$ where $(-b,c)$ is the vertex. Here, vertex is $(0,1)$, so $b=0$ and $c=1$. Thus: $$f(x)=a(x+0)^2+1 = a x^2 + 1$$ 4. **Finding $a$ using a point on the curve:** Use point $(2,-1)$: $$-1 = a(2)^2 + 1$$ $$-1 = 4a + 1$$ $$4a = -2$$ $$a = -\frac{1}{2}$$ 5. **Final form of $f(x)$:** $$f(x) = -\frac{1}{2} x^2 + 1$$ 6. **Check with point $(-3,14)$:** $$f(-3) = -\frac{1}{2}(-3)^2 + 1 = -\frac{1}{2} \times 9 + 1 = -4.5 + 1 = -3.5$$ This does not match $14$, so the point $(-3,14)$ is likely not on the curve or there is a misunderstanding. 7. **Summary:** (a) Axis of symmetry: $x=0$ (b) $f(x) = -\frac{1}{2} x^2 + 1$ which is in the form $(x+0)^2 + 1$ with $a=-\frac{1}{2}$, $b=0$, $c=1$. **Note:** The tangent line $y=-1$ touches the curve at $x=2$ since $f(2)=-1$ and the derivative at $x=2$ equals 0 (confirming tangency).