Given the function $f(x) = -x^2 + 6x - 11$, we need to find: (a) $f(2)$ (b) $f(-10)$ (c) $f(t)$ (d) $f(t-3)$ (e) $f(x-3)$ (f) $f(4x-1)$

1. Problem Description

Given the function $f(x) = -x^2 + 6x - 11$, we need to find:

(a) $f(2)$

(b) $f(-10)$

(c) $f(t)$

(d) $f(t-3)$

(e) $f(x-3)$

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

2. Solution Steps

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

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

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

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

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

$= -t^2 + 6t - 9 + 6t - 18 - 11$

$= -t^2 + 12t - 38$

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

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

$= -x^2 + 6x - 9 + 6x - 18 - 11$

$= -x^2 + 12x - 38$

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

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

$= -16x^2 + 8x - 1 + 24x - 6 - 11$

$= -16x^2 + 32x - 18$

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