1. The problem is to find a quadratic function $y = ax^2 + bx + c$ given the y-intercept and roots. 2. The y-intercept is the point where $x=0$, so $y=c$. Given the y-intercept $(0,-1)$, we have $c = -1$. 3. The roots are the values of $x$ where $y=0$. Given roots $-1$ and $4$, the quadratic can be expressed as $y = a(x + 1)(x - 4)$. 4. Substitute $x=0$ and $y=-1$ into the factored form to find $a$: $$-1 = a(0 + 1)(0 - 4) = a(1)(-4) = -4a$$ 5. Solve for $a$: $$-1 = -4a \implies a = \frac{1}{4}$$ 6. Therefore, the quadratic function is: $$y = \frac{1}{4}(x + 1)(x - 4)$$ 7. Expanding the expression: $$y = \frac{1}{4}(x^2 - 3x - 4) = \frac{1}{4}x^2 - \frac{3}{4}x - 1$$ Final answer: $$y = \frac{1}{4}x^2 - \frac{3}{4}x - 1$$