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$1.4 Pre cal-Quadratic formula 7. The profits of Mr. Unlucky's company can be represented by the equation p=-3t^(2)+18 t-4, where p is the amount of profit in hundreds of thousands of dollars and x is the number of years of operation. He realizes his company is on the downturn and wishes to sell before he ends up in debt. [6 pts] a) When will Unlucky's business show the maximum profit? {:[P=-3t^(2)+18 t-4,P+4-27=-3(t^(2)-6t)],[P+4=-3t^(2)+18 t,P-23=-3(t^(2)-6t:}],[P+4=-3(t^(2)-6t)rarr-6-7-3rarr i,P=-3(t-3)^(2)+23" unluchy's boisne show the max' "],[," vertex "=(3)23)" Profit after "]:} b) At what time will it be too late to sell his business? (When will he start losing money?)$

$1$ . $4$ Pre cal-Quadratic formula $\newline$ $7$ . The profits of Mr. Unlucky's company can be represented by the equation $p=-3 t^{2}+18 t-4$ , where $p$ is the amount of profit in hundreds of thousands of dollars and $x$ is the number of years of operation. He realizes his company is on the downturn and wishes to sell before he ends up in debt. [ $6$ pts] $\newline$ a) When will Unlucky's business show the maximum profit? $\newline$ $\begin{array}{ll} P=-3 t^{2}+18 t-4 & P+4-27=-3\left(t^{2}-6 t\right) \\ P+4=-3 t^{2}+18 t & P-23=-3\left(t^{2}-6 t\right. \\ P+4=-3\left(t^{2}-6 t\right) \rightarrow-6-7-3 \rightarrow i & P=-3(t-3)^{2}+23 \text { unluchy's boisne show the max' } \\ & \text { vertex }=(3) 23) \text { Profit after } \end{array}$ $\newline$ b) At what time will it be too late to sell his business? (When will he start losing money?)

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Solve linear equations with variables on both sides: word problems

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1

4

Pre cal-Quadratic formula

\newline

7

. The profits of Mr. Unlucky's company can be represented by the equation

p=-3 t^{2}+18 t-4

, where

p

is the amount of profit in hundreds of thousands of dollars and

x

is the number of years of operation. He realizes his company is on the downturn and wishes to sell before he ends up in debt. [

6

pts]

\newline

a) When will Unlucky's business show the maximum profit?

\newline

\begin{array}{ll} P=-3 t^{2}+18 t-4 & P+4-27=-3\left(t^{2}-6 t\right) \\ P+4=-3 t^{2}+18 t & P-23=-3\left(t^{2}-6 t\right. \\ P+4=-3\left(t^{2}-6 t\right) \rightarrow-6-7-3 \rightarrow i & P=-3(t-3)^{2}+23 \text { unluchy's boisne show the max' } \\ & \text { vertex }=(3) 23) \text { Profit after } \end{array}

\newline

b) At what time will it be too late to sell his business? (When will he start losing money?)

Find Vertex of Parabola: To find the maximum profit, we need to find the vertex of the parabola represented by the profit equation $p = -3t^2 + 18t - 4$ .
Calculate Vertex Coordinates: The vertex form of a quadratic equation is $p = a(t-h)^2 + k$ , where $(h, k)$ is the vertex of the parabola.
Determine Maximum Profit: The $t$ -coordinate of the vertex ( $h$ ) is found using the formula $h = -\frac{b}{2a}$ . For the given equation, $a = -3$ and $b = 18$ .
Find When Company Loses Money: Calculate $h$ : $h = \frac{-18}{2 \cdot (-3)} = \frac{-18}{-6} = 3$ .
Find When Company Loses Money: Calculate $h$ : $h = \frac{-18}{2 \cdot (-3)} = \frac{-18}{-6} = 3$ .Now we need to find the $k$ value by substituting $t = 3$ into the original equation.
Find When Company Loses Money: Calculate $h$ : $h = -18/(2*(-3)) = -18/(-6) = 3$ . Now we need to find the $k$ value by substituting $t = 3$ into the original equation. Calculate $k$ : $p = -3(3)^2 + 18(3) - 4 = -3(9) + 54 - 4 = -27 + 54 - 4 = 23$ .
Find When Company Loses Money: Calculate $h$ : $h = -18/(2*(-3)) = -18/(-6) = 3$ .Now we need to find the $k$ value by substituting $t = 3$ into the original equation.Calculate $k$ : $p = -3(3)^2 + 18(3) - 4 = -3(9) + 54 - 4 = -27 + 54 - 4 = 23$ .The vertex of the parabola is $(3, 23)$ , which means the maximum profit occurs at $3$ years of operation.
Find When Company Loses Money: Calculate $h$ : $h = -18/(2*(-3)) = -18/(-6) = 3$ .Now we need to find the $k$ value by substituting $t = 3$ into the original equation.Calculate $k$ : $p = -3(3)^2 + 18(3) - 4 = -3(9) + 54 - 4 = -27 + 54 - 4 = 23$ .The vertex of the parabola is $(3, 23)$ , which means the maximum profit occurs at $3$ years of operation.To find when the company starts losing money, we need to determine when the profit $p$ becomes less than $0$ .
Find When Company Loses Money: Calculate $h$ : $h = -\frac{18}{2(-3)} = \frac{-18}{-6} = 3$ .Now we need to find the $k$ value by substituting $t = 3$ into the original equation.Calculate $k$ : $p = -3(3)^2 + 18(3) - 4 = -3(9) + 54 - 4 = -27 + 54 - 4 = 23$ .The vertex of the parabola is $(3, 23)$ , which means the maximum profit occurs at $3$ years of operation.To find when the company starts losing money, we need to determine when the profit $p$ becomes less than $0$ .Set the profit equation to zero and solve for $t$ : $h = -\frac{18}{2(-3)} = \frac{-18}{-6} = 3$ $0$ .
Find When Company Loses Money: Calculate $h$ : $h = -18/(2*(-3)) = -18/(-6) = 3$ .Now we need to find the $k$ value by substituting $t = 3$ into the original equation.Calculate $k$ : $p = -3(3)^2 + 18(3) - 4 = -3(9) + 54 - 4 = -27 + 54 - 4 = 23$ .The vertex of the parabola is $(3, 23)$ , which means the maximum profit occurs at $3$ years of operation.To find when the company starts losing money, we need to determine when the profit $p$ becomes less than $0$ .Set the profit equation to zero and solve for $h = -18/(2*(-3)) = -18/(-6) = 3$ $0$ : $h = -18/(2*(-3)) = -18/(-6) = 3$ $1$ .Use the quadratic formula $h = -18/(2*(-3)) = -18/(-6) = 3$ $2$ to find the values of $h = -18/(2*(-3)) = -18/(-6) = 3$ $0$ .
Find When Company Loses Money: Calculate $h$ : $h = -\frac{18}{2(-3)} = \frac{-18}{-6} = 3$ . Now we need to find the $k$ value by substituting $t = 3$ into the original equation. Calculate $k$ : $p = -3(3)^2 + 18(3) - 4 = -3(9) + 54 - 4 = -27 + 54 - 4 = 23$ . The vertex of the parabola is $(3, 23)$ , which means the maximum profit occurs at $3$ years of operation. To find when the company starts losing money, we need to determine when the profit $p$ becomes less than $0$ . Set the profit equation to zero and solve for $h = -\frac{18}{2(-3)} = \frac{-18}{-6} = 3$ $0$ : $h = -\frac{18}{2(-3)} = \frac{-18}{-6} = 3$ $1$ . Use the quadratic formula $h = -\frac{18}{2(-3)} = \frac{-18}{-6} = 3$ $2$ to find the values of $h = -\frac{18}{2(-3)} = \frac{-18}{-6} = 3$ $0$ . Substitute $h = -\frac{18}{2(-3)} = \frac{-18}{-6} = 3$ $4$ , $h = -\frac{18}{2(-3)} = \frac{-18}{-6} = 3$ $5$ , and $h = -\frac{18}{2(-3)} = \frac{-18}{-6} = 3$ $6$ into the quadratic formula.
Find When Company Loses Money: Calculate $h$ : $h = -18/(2*(-3)) = -18/(-6) = 3$ .Now we need to find the $k$ value by substituting $t = 3$ into the original equation.Calculate $k$ : $p = -3(3)^2 + 18(3) - 4 = -3(9) + 54 - 4 = -27 + 54 - 4 = 23$ .The vertex of the parabola is $(3, 23)$ , which means the maximum profit occurs at $3$ years of operation.To find when the company starts losing money, we need to determine when the profit $p$ becomes less than $0$ .Set the profit equation to zero and solve for $h = -18/(2*(-3)) = -18/(-6) = 3$ $0$ : $h = -18/(2*(-3)) = -18/(-6) = 3$ $1$ .Use the quadratic formula $h = -18/(2*(-3)) = -18/(-6) = 3$ $2$ to find the values of $h = -18/(2*(-3)) = -18/(-6) = 3$ $0$ .Substitute $h = -18/(2*(-3)) = -18/(-6) = 3$ $4$ , $h = -18/(2*(-3)) = -18/(-6) = 3$ $5$ , and $h = -18/(2*(-3)) = -18/(-6) = 3$ $6$ into the quadratic formula.Calculate the discriminant: $h = -18/(2*(-3)) = -18/(-6) = 3$ $7$ .
Find When Company Loses Money: Calculate $h$ : $h = -18/(2*(-3)) = -18/(-6) = 3$ .Now we need to find the $k$ value by substituting $t = 3$ into the original equation.Calculate $k$ : $p = -3(3)^2 + 18(3) - 4 = -3(9) + 54 - 4 = -27 + 54 - 4 = 23$ .The vertex of the parabola is $(3, 23)$ , which means the maximum profit occurs at $3$ years of operation.To find when the company starts losing money, we need to determine when the profit $p$ becomes less than $0$ .Set the profit equation to zero and solve for $h = -18/(2*(-3)) = -18/(-6) = 3$ $0$ : $h = -18/(2*(-3)) = -18/(-6) = 3$ $1$ .Use the quadratic formula $h = -18/(2*(-3)) = -18/(-6) = 3$ $2$ to find the values of $h = -18/(2*(-3)) = -18/(-6) = 3$ $0$ .Substitute $h = -18/(2*(-3)) = -18/(-6) = 3$ $4$ , $h = -18/(2*(-3)) = -18/(-6) = 3$ $5$ , and $h = -18/(2*(-3)) = -18/(-6) = 3$ $6$ into the quadratic formula.Calculate the discriminant: $h = -18/(2*(-3)) = -18/(-6) = 3$ $7$ .Calculate the two possible values for $h = -18/(2*(-3)) = -18/(-6) = 3$ $0$ : $h = -18/(2*(-3)) = -18/(-6) = 3$ $9$ .
Find When Company Loses Money: Calculate $h$ : $h = -18/(2*(-3)) = -18/(-6) = 3$ . Now we need to find the $k$ value by substituting $t = 3$ into the original equation. Calculate $k$ : $p = -3(3)^2 + 18(3) - 4 = -3(9) + 54 - 4 = -27 + 54 - 4 = 23$ . The vertex of the parabola is $(3, 23)$ , which means the maximum profit occurs at $3$ years of operation. To find when the company starts losing money, we need to determine when the profit $p$ becomes less than $0$ . Set the profit equation to zero and solve for $h = -18/(2*(-3)) = -18/(-6) = 3$ $0$ : $h = -18/(2*(-3)) = -18/(-6) = 3$ $1$ . Use the quadratic formula $h = -18/(2*(-3)) = -18/(-6) = 3$ $2$ to find the values of $h = -18/(2*(-3)) = -18/(-6) = 3$ $0$ . Substitute $h = -18/(2*(-3)) = -18/(-6) = 3$ $4$ , $h = -18/(2*(-3)) = -18/(-6) = 3$ $5$ , and $h = -18/(2*(-3)) = -18/(-6) = 3$ $6$ into the quadratic formula. Calculate the discriminant: $h = -18/(2*(-3)) = -18/(-6) = 3$ $7$ . Calculate the two possible values for $h = -18/(2*(-3)) = -18/(-6) = 3$ $0$ : $h = -18/(2*(-3)) = -18/(-6) = 3$ $9$ . Simplify the square root: $k$ $0$ .
Find When Company Loses Money: Calculate $h$ : $h = -\frac{18}{2(-3)} = \frac{-18}{-6} = 3$ . Now we need to find the $k$ value by substituting $t = 3$ into the original equation. Calculate $k$ : $p = -3(3)^2 + 18(3) - 4 = -3(9) + 54 - 4 = -27 + 54 - 4 = 23$ . The vertex of the parabola is $(3, 23)$ , which means the maximum profit occurs at $3$ years of operation. To find when the company starts losing money, we need to determine when the profit $p$ becomes less than $0$ . Set the profit equation to zero and solve for $h = -\frac{18}{2(-3)} = \frac{-18}{-6} = 3$ $0$ : $h = -\frac{18}{2(-3)} = \frac{-18}{-6} = 3$ $1$ . Use the quadratic formula $h = -\frac{18}{2(-3)} = \frac{-18}{-6} = 3$ $2$ to find the values of $h = -\frac{18}{2(-3)} = \frac{-18}{-6} = 3$ $0$ . Substitute $h = -\frac{18}{2(-3)} = \frac{-18}{-6} = 3$ $4$ , $h = -\frac{18}{2(-3)} = \frac{-18}{-6} = 3$ $5$ , and $h = -\frac{18}{2(-3)} = \frac{-18}{-6} = 3$ $6$ into the quadratic formula. Calculate the discriminant: $h = -\frac{18}{2(-3)} = \frac{-18}{-6} = 3$ $7$ . Calculate the two possible values for $h = -\frac{18}{2(-3)} = \frac{-18}{-6} = 3$ $0$ : $h = -\frac{18}{2(-3)} = \frac{-18}{-6} = 3$ $9$ . Simplify the square root: $k$ $0$ . Now calculate the two values for $h = -\frac{18}{2(-3)} = \frac{-18}{-6} = 3$ $0$ : $k$ $2$ .
Find When Company Loses Money: Calculate $h$ : $h = -18/(2*(-3)) = -18/(-6) = 3$ .Now we need to find the $k$ value by substituting $t = 3$ into the original equation.Calculate $k$ : $p = -3(3)^2 + 18(3) - 4 = -3(9) + 54 - 4 = -27 + 54 - 4 = 23$ .The vertex of the parabola is $(3, 23)$ , which means the maximum profit occurs at $3$ years of operation.To find when the company starts losing money, we need to determine when the profit $p$ becomes less than $0$ .Set the profit equation to zero and solve for $h = -18/(2*(-3)) = -18/(-6) = 3$ $0$ : $h = -18/(2*(-3)) = -18/(-6) = 3$ $1$ .Use the quadratic formula $h = -18/(2*(-3)) = -18/(-6) = 3$ $2$ to find the values of $h = -18/(2*(-3)) = -18/(-6) = 3$ $0$ .Substitute $h = -18/(2*(-3)) = -18/(-6) = 3$ $4$ , $h = -18/(2*(-3)) = -18/(-6) = 3$ $5$ , and $h = -18/(2*(-3)) = -18/(-6) = 3$ $6$ into the quadratic formula.Calculate the discriminant: $h = -18/(2*(-3)) = -18/(-6) = 3$ $7$ .Calculate the two possible values for $h = -18/(2*(-3)) = -18/(-6) = 3$ $0$ : $h = -18/(2*(-3)) = -18/(-6) = 3$ $9$ .Simplify the square root: $k$ $0$ .Now calculate the two values for $h = -18/(2*(-3)) = -18/(-6) = 3$ $0$ : $k$ $2$ .Simplify the expression: $k$ $3$ .
Find When Company Loses Money: Calculate $h$ : $h = -18/(2*(-3)) = -18/(-6) = 3$ .Now we need to find the $k$ value by substituting $t = 3$ into the original equation.Calculate $k$ : $p = -3(3)^2 + 18(3) - 4 = -3(9) + 54 - 4 = -27 + 54 - 4 = 23$ .The vertex of the parabola is $(3, 23)$ , which means the maximum profit occurs at $3$ years of operation.To find when the company starts losing money, we need to determine when the profit $p$ becomes less than $0$ .Set the profit equation to zero and solve for $h = -18/(2*(-3)) = -18/(-6) = 3$ $0$ : $h = -18/(2*(-3)) = -18/(-6) = 3$ $1$ .Use the quadratic formula $h = -18/(2*(-3)) = -18/(-6) = 3$ $2$ to find the values of $h = -18/(2*(-3)) = -18/(-6) = 3$ $0$ .Substitute $h = -18/(2*(-3)) = -18/(-6) = 3$ $4$ , $h = -18/(2*(-3)) = -18/(-6) = 3$ $5$ , and $h = -18/(2*(-3)) = -18/(-6) = 3$ $6$ into the quadratic formula.Calculate the discriminant: $h = -18/(2*(-3)) = -18/(-6) = 3$ $7$ .Calculate the two possible values for $h = -18/(2*(-3)) = -18/(-6) = 3$ $0$ : $h = -18/(2*(-3)) = -18/(-6) = 3$ $9$ .Simplify the square root: $k$ $0$ .Now calculate the two values for $h = -18/(2*(-3)) = -18/(-6) = 3$ $0$ : $k$ $2$ .Simplify the expression: $k$ $3$ .The two values of $h = -18/(2*(-3)) = -18/(-6) = 3$ $0$ are $k$ $5$ and $k$ $6$ . Since we are looking for when the company starts losing money, we take the larger value.
Find When Company Loses Money: Calculate $h$ : $h = -18/(2*(-3)) = -18/(-6) = 3$ .Now we need to find the $k$ value by substituting $t = 3$ into the original equation.Calculate $k$ : $p = -3(3)^2 + 18(3) - 4 = -3(9) + 54 - 4 = -27 + 54 - 4 = 23$ .The vertex of the parabola is $(3, 23)$ , which means the maximum profit occurs at $3$ years of operation.To find when the company starts losing money, we need to determine when the profit $p$ becomes less than $0$ .Set the profit equation to zero and solve for $h = -18/(2*(-3)) = -18/(-6) = 3$ $0$ : $h = -18/(2*(-3)) = -18/(-6) = 3$ $1$ .Use the quadratic formula $h = -18/(2*(-3)) = -18/(-6) = 3$ $2$ to find the values of $h = -18/(2*(-3)) = -18/(-6) = 3$ $0$ .Substitute $h = -18/(2*(-3)) = -18/(-6) = 3$ $4$ , $h = -18/(2*(-3)) = -18/(-6) = 3$ $5$ , and $h = -18/(2*(-3)) = -18/(-6) = 3$ $6$ into the quadratic formula.Calculate the discriminant: $h = -18/(2*(-3)) = -18/(-6) = 3$ $7$ .Calculate the two possible values for $h = -18/(2*(-3)) = -18/(-6) = 3$ $0$ : $h = -18/(2*(-3)) = -18/(-6) = 3$ $9$ .Simplify the square root: $k$ $0$ .Now calculate the two values for $h = -18/(2*(-3)) = -18/(-6) = 3$ $0$ : $k$ $2$ .Simplify the expression: $k$ $3$ .The two values of $h = -18/(2*(-3)) = -18/(-6) = 3$ $0$ are $k$ $5$ and $k$ $6$ . Since we are looking for when the company starts losing money, we take the larger value.Calculate the larger value: $k$ $5$ . This is the time after which the company will start losing money.