Suppose that the function $f$ is defined, for all real numbers, as follows. \[ f(x)=\left\{\begin{array}{ll} \frac{1}{4} x^{2}-5 & \text { if } x \neq 1 \\ -2 & \text { if } x=1 \end{array}\right. \] Find $f(-1), f(1)$, and $f(4)$. \[ \begin{array}{c} f(-1)= \\ f(1)= \\ f(4)= \end{array} \]

Step 1 ：Given the function $f(x)$ defined as follows: $f(x)=\left\{\begin{array}{ll} \frac{1}{4} x^{2}-5 & \text { if } x \neq 1 \\ -2 & \text { if } x=1 \end{array}\right.$

Step 2 ：We need to find the values of $f(-1)$, $f(1)$, and $f(4)$.

Step 3 ：For $f(-1)$, since $-1 \neq 1$, we use the first part of the function definition. Substituting $x$ with $-1$ gives us $f(-1) = \frac{1}{4}(-1)^{2}-5 = -4.75$

Step 4 ：For $f(1)$, since $x = 1$, we use the second part of the function definition. So, $f(1) = -2$

Step 5 ：For $f(4)$, since $4 \neq 1$, we use the first part of the function definition. Substituting $x$ with $4$ gives us $f(4) = \frac{1}{4}(4)^{2}-5 = -1.0$

Step 6 ：So, the final answers are: $f(-1)= \boxed{-4.75}$, $f(1)= \boxed{-2}$, and $f(4)= \boxed{-1.0}$