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# Evaluate Polynomials

For ACT Students
The ACT is a timed exam...$60$ minutes for $60$ questions
So, you should try to solve each question correctly and timely
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no "negative" penalty for any wrong answer.

Evaluate these polynomial functions for each value of the independent variable.
Show all work.

(1.) $g(x) = 3x^2 - 3x + 1$

Calculate $g(0),\:\:\: g(-2),\:\:\: g(3),\:\:\: g(-x),\:\:\: g(1 - k)$

$g(0)$
This means that we have to substitute $0$ for $x$

$g(x) = 3x^2 - 3x + 1 \\[3ex] g(0) = 3 * 0^2 - 3 * 0 + 1 \\[3ex] = 3(0) - 0 + 1 \\[3ex] = 0 - 0 + 1 \\[3ex] = 1 \\[3ex]$ $g(-2)$
This means that we have to substitute $-2$ for $x$

$g(x) = 3x^2 - 3x + 1 \\[3ex] g(-2) = 3 * (-2)^2 - 3 * (-2) + 1 \\[3ex] = 3(4) + 6 + 1 \\[3ex] = 12 + 6 + 1 \\[3ex] = 19 \\[3ex]$ $g(3)$
This means that we have to substitute $3$ for $x$

$g(x) = 3x^2 - 3x + 1 \\[3ex] g(3) = 3 * (3)^2 - 3 * (3) + 1 \\[3ex] = 3(9) - 9 + 1 \\[3ex] = 27 - 9 + 1 \\[3ex] = 19 \\[3ex]$ $g(-x)$
This means that we have to substitute $-x$ for $x$

$g(x) = 3x^2 - 3x + 1 \\[3ex] g(-x) = 3 * (-x)^2 - 3 * (-x) + 1 \\[3ex] = 3 * x^2 + 3x + 1 \\[3ex] = 3x^2 + 3x + 1 \\[3ex]$ $g(1-k)$
This means that we have to substitute $(1-k)$ for $x$

$g(x) = 3x^2 - 3x + 1 \\[3ex] g(1-k) = 3 * (1-k)^2 - 3 * (1-k) + 1 \\[3ex] = 3 * (1-k) * (1-k) - 3(1-k) + 1 \\[3ex] = 3[(1-k)(1-k)] - 3 + 3k + 1 \\[3ex] = 3(1 - k - k + k^2) + 3k - 2 \\[3ex] = 3(1 - 2k + k^2) + 3k - 2 \\[3ex] = 3 - 6k + 3k^2 + 3k - 2 \\[3ex] = 3k^2 - 3k + 1$
(2.) $f(x) = -3x + 7$

Calculate $f(3),\:\:\: f(-5)$

$f(3)$
This means that we have to substitute $3$ for $x$

$f(x) = -3x + 7 \\[3ex] f(3) = -3(3) + 7 \\[3ex] f(3) = -9 + 7 \\[3ex] f(3) = -2 \\[3ex]$ $f(-5)$
This means that we have to substitute $-5$ for $x$

$f(x) = -3x + 7 \\[3ex] f(-5) = -3(-5) + 7 \\[3ex] f(-5) = 15 + 7 \\[3ex] f(-5) = 22$
(3.) $g(x) = {\dfrac{2}{3}}x - \dfrac{3}{4}$

Calculate $g(3),\:\:\: g(4),\:\:\: g\left(-\dfrac{1}{2}\right)$

$g(3)$
This means that we have to substitute $3$ for $x$

$g(x) = \dfrac{2}{3}x - \dfrac{3}{4} \\[5ex] g(x) = \dfrac{2}{3} * x - \dfrac{3}{4} \\[5ex] g(3) = \dfrac{2}{3} * 3 - \dfrac{3}{4} \\[5ex] = \dfrac{2}{1} - \dfrac{3}{4} \\[5ex] = \dfrac{8}{4} - \dfrac{3}{4} \\[5ex] = \dfrac{8 - 3}{4} \\[5ex] = \dfrac{5}{4} \\[5ex]$ $g(4)$
This means that we have to substitute $4$ for $x$

$g(x) = \dfrac{2}{3}x - \dfrac{3}{4} \\[5ex] g(x) = \dfrac{2}{3} * x - \dfrac{3}{4} \\[5ex] g(3) = \dfrac{2}{3} * 4 - \dfrac{3}{4} \\[5ex] = \dfrac{8}{4} - \dfrac{3}{4} \\[5ex] = \dfrac{32}{12} - \dfrac{9}{12} \\[5ex] = \dfrac{32 - 9}{12} \\[5ex] = \dfrac{23}{12} \\[5ex]$ $g\left(-\dfrac{1}{2}\right)$
This means that we have to substitute $-\dfrac{1}{2}$ for $x$

$g(x) = \dfrac{2}{3}x - \dfrac{3}{4} \\[5ex] g(x) = \dfrac{2}{3} * x - \dfrac{3}{4} \\[5ex] g(4) = \dfrac{2}{3} * -\dfrac{1}{2} - \dfrac{3}{4} \\[5ex] = -\dfrac{1}{3} - \dfrac{3}{4} \\[5ex] = -\dfrac{4}{12} - \dfrac{9}{12} \\[5ex] = \dfrac{-4 - 9}{12} \\[5ex] = -\dfrac{13}{12}$
(4.) $f(x) = x^2 - 3x$

Calculate $f(-a),\:\:\: f(a - 3),\:\:\: f(a + h)$

$f(-a)$
This means that we have to substitute $(-a)$ for $x$

$f(x) = x^2 - 3x \\[3ex] f(-a) = (-a)^2 - 3(-a) \\[3ex] = a^2 + 3a \\[3ex] = a(a + 3) \\[3ex]$ $f(a - 3)$
This means that we have to substitute $(a - 3)$ for $x$

$f(x) = x^2 - 3x \\[3ex] f(a - 3) = (a - 3)^2 - 3(a - 3) \\[3ex] = (a - 3)(a - 3) - 3a + 9 \\[3ex] = a^2 - 3a - 3a + 9 - 3a + 9 \\[3ex] = a^2 - 9a + 18 \\[3ex] = (a - 3)(a - 6) \\[3ex]$ OR

$f(x) = x^2 - 3x \\[3ex] f(a - 3) = (a - 3)^2 - 3(a - 3) \\[3ex] = (a - 3)(a - 3) - 3a + 9 \\[3ex] = (a - 3)(a - 3) - 3(a - 3) \\[3ex] = (a - 3)[(a - 3) - 3] \\[3ex] = (a - 3)[a - 3 - 3] \\[3ex] = (a - 3)(a -6) \\[3ex]$ $f(a + h)$
This means that we have to substitute $(a + h)$ for $x$

$f(x) = x^2 - 3x \\[3ex] f(a + h) = (a + h)^2 - 3(a + h) \\[3ex] = (a + h)(a + h) - 3(a + h) \\[3ex] = (a + h)[(a + h) - 3] \\[3ex] = (a + h)(a + h - 3)$
(5.) $p(m) = m^3$

Calculate $p(3),\:\:\: p(-2),\:\:\: p(-m),\:\:\: p(3y),\:\:\: p(2 + h)$

$p(3)$
This means that we have to substitute $3$ for $m$

$p(m) = m^3 \\[3ex] p(3) = 3^3 \\[3ex] = 27 \\[3ex]$ $p(-2)$
This means that we have to substitute $(-2)$ for $m$

$p(m) = m^3 \\[3ex] p(-2) = (-2)^3 \\[3ex] = -8 \\[3ex]$ $p(-m)$
This means that we have to substitute $(-m)$ for $m$

$p(m) = m^3 \\[3ex] p(3) = (-m)^3 \\[3ex] = (-m)(-m)(-m) \\[3ex] = -m^3 \\[3ex]$ $p(3y)$
This means that we have to substitute $(3y)$ for $m$

$p(m) = m^3 \\[3ex] p(3y) = (3y)^3 \\[3ex] = 27y^3 \\[3ex]$ $p(2 + h)$
This means that we have to substitute $(2 + h)$ for $m$

$p(m) = m^3 \\[3ex] p(2 + h) = (2 + h)^3 \\[3ex] = (2 + h)(2 + h)(2 + h) \\[3ex] = (4 + 2h + 2h + h^2)(2 + h) \\[3ex] = (h^2 + 4h + 4)(2 + h) \\[3ex] = (2h^2 + h^3 + 8h + 4h^2 + 8 + 4h) \\[3ex] = h^3 + 6h^2 + 12h + 8$
(6.) $p(x) = -x^2 - 5x + 8$

Calculate $p(0),\:\:\: p(7),\:\:\: p(-7),\:\:\: p(g)$

$p(0)$
This means that we have to substitute $0$ for $x$

$p(x) = -x^2 - 5x + 8 \\[3ex] p(x) = -1 * x^2 - 5x + 8 \\[3ex] p(0) = -1 * 0^2 - 5(0) + 8 \\[3ex] p(0) = -1 * 0 - 0 + 8 \\[3ex] p(0) = 0 - 0 + 8 \\[3ex] p(0) = 8 \\[3ex]$ $p(7)$
This means that we have to substitute $7$ for $x$

$p(x) = -x^2 - 5x + 8 \\[3ex] p(x) = -1 * x^2 - 5x + 8 \\[3ex] p(7) = -1 * 7^2 - 5(7) + 8 \\[3ex] p(7) = -1 * 49 - 35 + 8 \\[3ex] p(7) = -49 - 35 + 8 \\[3ex] p(7) = -76 \\[3ex]$ $p(-7)$
This means that we have to substitute $-7$ for $x$

$p(x) = -x^2 - 5x + 8 \\[3ex] p(x) = -1 * x^2 - 5x + 8 \\[3ex] p(-7) = -1 * (-7)^2 - 5(-7) + 8 \\[3ex] p(-7) = -1 * 49 + 35 + 8 \\[3ex] p(-7) = -49 + 35 + 8 \\[3ex] p(-7) = -6 \\[3ex]$ $p(g)$
This means that we have to substitute $g$ for $x$

$p(x) = -x^2 - 5x + 8 \\[3ex] p(g) = -g^2 - 5g + 8$
(7.) $f(x) = -4x + 7$

Calculate $f(a),\:\:\: f(a + 2),\:\:\: f(a + h)$

$f(a)$
This means that we have to substitute $a$ for $x$

$f(x) = -4x + 7 \\[3ex] f(a) = -4a + 7 \\[3ex]$
$f(a + 2)$
This means that we have to substitute $(a + 2)$ for $x$

$f(x) = -4x + 7 \\[3ex] f(a + 2) = -4(a + 2) + 7 \\[3ex] = -4a - 8 + 7 \\[3ex] = -4a - 1 \\[3ex]$
$f(a + h)$
This means that we have to substitute $(a + h)$ for $x$

$f(x) = -4x + 7 \\[3ex] f(a + h) = -4(a + h) + 7 \\[3ex] = -4a - 4h + 7$
(8.) ACT The function $f$ is defined as $f(x) = -4x^3 - 4x^2$.
What is $f(-4)$

$F.\:\: -320 \\[3ex] G.\:\: -192 \\[3ex] H.\:\: 16 \\[3ex] J.\:\: 192 \\[3ex] K.\:\: 320 \\[3ex]$

$f(x) = -4x^3 - 4x^2 \\[3ex] f(-4) \rightarrow x = -4 \\[3ex] f(-4) = -4[(-4)^3] - 4(-4)^2 \\[3ex] f(-4) = -4(-64) - 4(16) \\[3ex] f(-4) = 256 - 64 \\[3ex] f(-4) = 192$