We are given the function $f(x) = -x^2 + 6x - 11$. We need to find the values of $f(2)$, $f(-10)$, $f(t)$, $f(t-3)$, $f(x-3)$, and $f(4x-1)$.

1. Problem Description

We are given the function $f(x) = -x^2 + 6x - 11$. We need to find the values of $f(2)$, $f(-10)$, $f(t)$, $f(t-3)$, $f(x-3)$, and $f(4x-1)$.

2. Solution Steps

(a) $f(2)$:

Substitute $x=2$ into the function:

$f(2) = -(2)^2 + 6(2) - 11 = -4 + 12 - 11 = 8 - 11 = -3$

(b) $f(-10)$:

Substitute $x=-10$ into the function:

$f(-10) = -(-10)^2 + 6(-10) - 11 = -100 - 60 - 11 = -171$

(c) $f(t)$:

Substitute $x=t$ into the function:

$f(t) = -t^2 + 6t - 11$

(d) $f(t-3)$:

Substitute $x=t-3$ into the function:

$f(t-3) = -(t-3)^2 + 6(t-3) - 11$

$f(t-3) = -(t^2 - 6t + 9) + 6t - 18 - 11$

$f(t-3) = -t^2 + 6t - 9 + 6t - 18 - 11$

$f(t-3) = -t^2 + 12t - 38$

(e) $f(x-3)$:

Substitute $x$ with $x-3$ in the function.

$f(x-3) = -(x-3)^2 + 6(x-3) - 11$

$f(x-3) = -(x^2 - 6x + 9) + 6x - 18 - 11$

$f(x-3) = -x^2 + 6x - 9 + 6x - 18 - 11$

$f(x-3) = -x^2 + 12x - 38$

(f) $f(4x-1)$:

Substitute $x=4x-1$ into the function:

$f(4x-1) = -(4x-1)^2 + 6(4x-1) - 11$

$f(4x-1) = -(16x^2 - 8x + 1) + 24x - 6 - 11$

$f(4x-1) = -16x^2 + 8x - 1 + 24x - 6 - 11$

$f(4x-1) = -16x^2 + 32x - 18$

3. Final Answer

(a) $f(2) = -3$

(b) $f(-10) = -171$

(c) $f(t) = -t^2 + 6t - 11$

(d) $f(t-3) = -t^2 + 12t - 38$

(e) $f(x-3) = -x^2 + 12x - 38$

(f) $f(4x-1) = -16x^2 + 32x - 18$