ancequation of a curve is $y=x^{2}-6 x+k$ , where $k$ is a constant. $\newline$ (i) Find the set of values of $k$ for which the whole of the curve lies above the $x$ -axis. $\newline$ (ii) Find the value of $k$ for which the line $y+2 x=7$ is a tangent to the curve. $\newline$ a company producing salt from sea water changed to a new process. The amount of sal $2 \%$ of the amount $\newline$ wh whe fler the change the company obtained obtained in the preceding week. It is given that $\newline$ reak after the change the company obtained $8000 \mathrm{~kg}$ of salt. $\newline$ (i) Find the amount of salt obtained in the $12$ th week after the change. $\newline$ (ii) Find the total amount of salt obtained in the first $12$ weeks after the change.

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Find a value using two-variable equations: word problems

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Q. ancequation of a curve is

y=x^{2}-6 x+k

, where

k

is a constant.

\newline

(i) Find the set of values of

k

for which the whole of the curve lies above the

x

-axis.

\newline

(ii) Find the value of

k

for which the line

y+2 x=7

is a tangent to the curve.

\newline

a company producing salt from sea water changed to a new process. The amount of sal

2 \%

of the amount

\newline

wh whe fler the change the company obtained obtained in the preceding week. It is given that

\newline

reak after the change the company obtained

8000 \mathrm{~kg}

of salt.

\newline

(i) Find the amount of salt obtained in the

12

th week after the change.

\newline

(ii) Find the total amount of salt obtained in the first

12

weeks after the change.

Identify Vertex: Identify the vertex of the parabola $y=x^{2}-6x+k$ to determine when it is above the x-axis. The vertex form of a parabola is $y=a(x-h)^{2}+k$ . Here, $a=1$ and the x-coordinate of the vertex, $h$ , is given by $-b/(2a) = 6/2 = 3$ .
Vertex Coordinates: Substitute $x=3$ into the equation to find the $y$ -coordinate of the vertex: $y=3^{2}-6\times 3+k = 9-18+k = -9+k$ . For the curve to be above the $x$ -axis, $-9+k > 0$ , hence $k > 9$ .
Tangent Condition: For the tangent condition, set the derivative of $y=x^{2}-6x+k$ equal to the slope of the line $y+2x=7$ , which simplifies to $y=-2x+7$ . The derivative, $\frac{dy}{dx} = 2x-6$ , must equal $-2$ .
Solve for $k$ : Solve $2x-6 = -2$ : $2x = 4$ , $x = 2$ . Substitute $x=2$ into the curve equation: $y=2^{2}-6\cdot2+k = 4-12+k = -8+k$ . Set this equal to $y$ value from the line at $x=2$ : $-8+k = -2x+7$ at $x=2$ , $2x-6 = -2$ $0$ .
Salt Production: Solve $-8+k = 3$ : $k = 11$ .
Calculate Final Amount: For the salt production, use the geometric series formula to find the amount in the $12^{th}$ week. The initial amount after the change is $8000\,\text{kg}$ , decreasing by $2\%$ each week. The amount in the $12^{th}$ week is $8000\times(0.98)^{11}$ .
Calculate Final Amount: For the salt production, use the geometric series formula to find the amount in the $12^{\text{th}}$ week. The initial amount after the change is $8000$ kg, decreasing by $2\%$ each week. The amount in the $12^{\text{th}}$ week is $8000 \times (0.98)^{11}$ . Calculate $8000 \times (0.98)^{11} \approx 8000 \times 0.886 = 7088$ kg.