(a)

\begin{align} xy & = 3x - 5 \\ \\ \therefore Y & = 3X - 5 \\ \\ \text{where } Y & = xy \\ X & = x \\ m & = 3 \\ c & = -5 \end{align}

(b)

\begin{align} y & = x^2 - 2x \\ {1 \over x}(y) & = {1 \over x}(x^2 - 2x) \\ {y \over x} & = x - 2 \\ \\ \therefore Y & = X - 2 \\ \\ \text{where } Y & = {y \over x} \\ X & = x \\ m & = 1 \\ c & = -2 \end{align}

(c)

\begin{align} y & = {3 \over x - 2} \\ y(x - 2) & = 3 \\ yx - 2y & = 3 \\ yx & = 2y + 3 \\ \\ \therefore Y & = 2X + 3 \\ \\ \text{where } Y & = yx \\ X & = y \\ m & = 2 \\ c & = 3 \end{align}

(c) Alternative solution:

\begin{align} y & = {3 \over x - 2} \\ y(x - 2) & = 3 \\ xy - 2y & = 3 \\ -2y & = 3 - xy \\ 2y & = xy - 3 \\ y & = {1 \over 2}xy - {3 \over 2} \\ \\ \therefore Y & = {1 \over 2}X - {3 \over 2} \\ \\ \text{where } Y & = y \\ X & = xy \\ m & = {1 \over 2} \\ c & = -{3 \over 2} \end{align}

(d)

\begin{align} 3y & = 2x - {5 \over x} \\ x(3y) & = x \left(2x - {5 \over x}\right) \\ 3xy & = 2x^2 - 5 \\ {1 \over 3}(3xy) & = {1 \over 3}(2x^2 - 5) \\ xy & = {2 \over 3}x^2 - {5 \over 3} \\ \\ \therefore Y & = {2 \over 3}X - {5 \over 3} \\ \\ \text{where } Y & = xy \\ X & = x^2 \\ m & = {2 \over 3} \\ c & = -{5 \over 3} \end{align}

(e)

\begin{align} y & = 10 \times 2^x \\ \lg y & = \lg (10 \times 2^x) \\ \lg y & = \lg 10 + \lg 2^x \phantom{00000} [\text{Product law (logarithms)}] \\ \lg y & = 1 + \lg 2^x \\ \lg y & = 1 + x\lg 2 \phantom{000000.0} [\text{Power law (logarithms)}] \\ \lg y & = (\lg 2)x + 1 \\ \\ \therefore Y & = (\lg 2) X + 1 \\ \\ \text{where } Y & = \lg y \\ X & = x \\ m & = \lg 2 \\ c & = 1 \end{align}

(f)

\begin{align} y & = 5x^7 \\ \lg y & = \lg (5x^7) \\ \lg y & = \lg 5 + \lg x^7 \phantom{00000} [\text{Product law (logarithms)}] \\ \lg y & = \lg 5 + 7\lg x \phantom{000/0} [\text{Power law (logarithms)}] \\ \lg y & = 7\lg x + \lg 5 \\ \\ \therefore Y & = 7X + \lg 5 \\ \\ \text{where } Y & = \lg y \\ X & = \lg x \\ m & = 7 \\ c & = \lg 5 \end{align}

(a)

\begin{align} y^2 & = 2x + 3 \\ \\ \therefore Y & = 2X + 3 \\ \\ \text{where } Y & = y^2 \\ X & = x \\ m & = 2 \\ c & = 3 \\ \\ \\ \text{When } & X = 0, \\ Y & = 2(0) + 3 \\ & = 3 \\ \\ \text{When } & Y = 0, \\ 0 & = 2X + 3 \\ -3 & = 2X \\ -{3 \over 2} & = X \end{align}
Thus, the line passes through the points $(0, 3)$ and $\left(-{3 \over 2}, 0 \right)$. Plotting $y^2$ against $x$:

(b)

\begin{align} xy & = 2y^2 - 5 \\ \\ \therefore Y & =2X - 5 \\ \\ \text{where } Y & = xy \\ X & = y^2 \\ m & = 2 \\ c & = -5 \\ \\ \\ \text{When } & X = 0, \\ Y & = 2(0) - 5 \\ & = -5 \\ \\ \text{When } & Y = 0, \\ 0 & = 2X - 5 \\ 5 & = 2X \\ {5 \over 2} & = X \end{align}
Thus, the line passes through the points $(0, -5)$ and $\left( {5 \over 2}, 0 \right)$. Plotting $xy$ against $y^2$:

(a)

\begin{align} xy^2 & = 2x + 5y \\ {1 \over x}(xy^2) & = {1 \over x}(2x + 5y) \\ y^2 & = 2 + 5\left({y \over x}\right) \\ y^2 & = 5\left({y \over x}\right) + 2 \\ \\ \therefore Y & = 5X + 2 \\ \\ \text{where } Y & = y^2 \\ X & = {y \over x} \\ m & = 5 \\ c & = 2 \end{align}

(b)

\begin{align} 3xy & = 5y - 2x \\ {1 \over y}(3xy) & = {1 \over y}(5y - 2x) \\ 3x & = 5 - {2x \over y} \\ {1 \over x}(3x) & = {1 \over x} \left(5 - {2x \over y}\right) \\ 3 & = {5 \over x} - {2 \over y} \\ {2 \over y} & = {5 \over x} - 3 \\ {1 \over y} & = {5 \over 2x} - {3 \over 2} \\ {1 \over y} & = {5 \over 2} \left(1 \over x\right) - {3 \over 2} \\ \\ \therefore Y & = {5 \over 2} X - {3 \over 2} \\ \\ \text{where } Y & = {1 \over y} \\ X & = {1 \over x} \\ m & = {5 \over 2} \\ c & = -{3 \over 2} \end{align}

(i)

\begin{align} y & = 5x^{3 \over 2} \\ \\ \text{Take } \lg & \text{ of both sides,} \\ \\ \lg y & = \lg (5x^{3 \over 2}) \\ \\ \lg y & = \lg 5 + \lg x^{3 \over 2} \phantom{00000} [\text{Product law (logarithms)}] \\ \\ \lg y & = \lg 5 + {3 \over 2}\lg x \phantom{0000/} [\text{Power law (logarithms)}] \\ \\ \lg y & = {3 \over 2}\lg x + \lg 5 \text{ (Shown)} \end{align}

(ii)

\begin{align} \lg y & = {3 \over 2}\lg x + \lg 5 \\ \\ \therefore Y & = {3 \over 2}X + \lg 5 \\ \\ \\ \text{When } & X = 0, \\ Y & = {3 \over 2}(0) + \lg 5 \\ & = \lg 5 \\ \\ \text{When } & Y = 0, \\ 0 & = {3 \over 2}X + \lg 5 \\ -\lg 5 & = {3 \over 2}X \\ \\ 3X & = -2\lg 5 \\ X & = -{2 \over 3}\lg 5 \end{align}
Thus, the line passes through the points $(0, \lg 5)$ and $\left(-{2 \over 3}\lg 5, 0\right)$. Plotting $\lg y$ against $\lg x$:

(a)

The line passes through the points (4, 9) and (0, 1) \begin{align} \text{Since } y & \text{-intercept is } 1, \\ c & = 1 \\ \\ m & = {9 - 1 \over 4 - 0} \\ & = 2 \\ \\ \text{Linear equation: } \phantom{0} & Y = 2X + 1 \\ \\ \\ \text{Since } Y = y & \text{ and } X = x^2, \\ \\ \text{Non-linear equation: } \phantom{0} & y = 2x^2 + 1 \end{align}

(b)

The line passes through the points (-4, 5) and (0, 1) \begin{align} \text{Since } y & \text{-intercept is } 1, \\ c & = 1 \\ \\ m & = {5 - 1 \over -4 - 0} \\ & = -1 \\ \\ \text{Linear equation: } \phantom{0} & Y = -X + 1 \\ \\ \\ \text{Since } Y = xy & \text{ and } X = x^3, \\ \\ xy & = -x^3 + 1 \\ {1 \over x}(xy) & = {1 \over x}(-x^3 + 1) \\ y & = -x^2 + {1 \over x} \\ \\ \text{Non-linear equation: } \phantom{0} & y = -x^2 + {1 \over x} \end{align}

(c)

The line passes through the points (0, 2) and (4, 0) \begin{align} \text{Since } y & \text{-intercept is } 2, \\ c & = 2 \\ \\ m & = {2 - 0 \over 0 - 4} \\ & = -{1 \over 2} \\ \\ \text{Linear equation: } \phantom{0} & Y = -{1 \over 2}X + 2 \\ \\ \\ \text{Since } Y = {y \over x} & \text{ and } X = x + y, \\ \\ {y \over x} & = -{1 \over 2}(x + y) + 2 \\ {y \over x} & = -{1 \over 2}x - {1 \over 2}y + 2 \\ y & = -{1 \over 2}x^2 - {1 \over 2}xy + 2x \\ 2y & = -x^2 - xy + 4x \\ 2y + xy & = 4x - x^2 \\ y(2 + x) & = 4x - x^2 \\ y & = {4x - x^2 \over 2 + x} \\ \\ \text{Non-linear equation: } \phantom{0} & y = {x(4 - x) \over x + 2} \end{align}

(d)

The line passes through the points (0, -2) and (4, 1) \begin{align} \text{Since } y & \text{-intercept is } - 2, \\ c & = -2 \\ \\ m & = {-2 - 1 \over 0 - 4} \\ & = {3 \over 4} \\ \\ \text{Linear equation: } \phantom{0} & Y = {3 \over 4}X - 2 \\ \\ \\ \text{Since } Y = {y \over x} & \text{ and } X = x, \\ \\ {y \over x} & = {3 \over 4}x - 2 \\ y & = x \left({3 \over 4}x - 2\right) \\ y & = {3 \over 4}x^2 - 2x \\ \\ \text{Non-linear equation: } \phantom{0} & y = {3 \over 4}x^2 - 2x \end{align}

(a)

\begin{align} m & = {3 - 2 \over 1 - 2} \\ & = -1 \\ \\ Y & = -X + c \\ \\ \text{When } X & = 1 \text{ and } Y = 3, \\ 3 & = -(1) + c \\ 3 + 1 & = c \\ 4 & = c \\ \\ \therefore \text{Linear equation: } \phantom{0} & Y = -X + 4 \\ \\ \\ \text{Since } Y = \lg y & \text{ and } X = \lg x, \\ \\ \lg y & = -\lg x + 4 \\ \lg y + \lg x & = 4 \\ \lg(y \times x) & = 4 \phantom{0000000000} [\text{Product law (logarithms)}] \\ \lg (xy) & = 4 \\ \log_{10} xy & = 4 \\ \\ xy & = 10^4 \phantom{00000000} [\text{Change to index form}] \\ xy & = 10000 \\ y & = {10000 \over x} \end{align}

(b)

\begin{align} m & = {4 - 3 \over 3 - 1} \\ & = {1 \over 2} \\ \\ Y & = {1 \over 2}X + c \\ \\ \text{When } X & = 1 \text{ and } Y = 3, \\ 3 & = {1 \over 2}(1) + c \\ 3 & = {1 \over 2} + c \\ 3 - {1 \over 2} & = c \\ {5 \over 2} & = c \\ \\ \therefore \text{Linear equation: } \phantom{0} & Y = {1 \over 2}X + {5 \over 2} \\ \\ \\ \text{Since } Y = {x + y \over x} & \text{ and } X = x, \\ \\ {x + y \over x} & = {1 \over 2}x + {5 \over 2} \\ x + y & = x \left( {1 \over 2}x + {5 \over 2} \right) \\ x + y & = {1 \over 2} x^2 + {5 \over 2}x \\ y & = {1 \over 2}x^2 + {5 \over 2}x - x \\ y & = {1 \over 2}x^2 + {3 \over 2}x \end{align}

(i)

\begin{align} y & = pq^x \\ \\ \text{Take } \ln & \text{ of both sides,} \\ \ln y & = \ln (pq^x) \\ \ln y & = \ln p + \ln q^x \phantom{00000000} [\text{Product law (logarithms)}] \\ \ln y & = \ln p + x\ln q \phantom{0000000/} [\text{Power law (logarithms)}] \\ \ln y & = (\ln q) x + \ln p \\ \\ \therefore Y & = (\ln q) X + \ln p \\ \\ \text{where } Y & = \ln y \\ X & = x \\ m & = \ln q \\ c & = \ln p \end{align}
From the question, the line has a gradient of $-{3 \over 2}$
\begin{align} m & = \ln q \\ -{3 \over 2} & = \ln q \\ -{3 \over 2} & = \log_e q \\ \\ e^{-{3 \over 2}} & = q \phantom{000000} [\text{Change to index form}] \\ \\ q & = 0.22313 \\ & \approx 0.2 \text{ (1 d.p.)} \end{align}
From the graph, the vertical intercept of the graph is 2
\begin{align} c & = \ln p \\ 2 & = \ln p \\ 2 & = \log_e p \\ \\ e^2 & = p \phantom{000000} [\text{Change to index form}] \\ \\ p & = 7.38905 \\ & \approx 7.4 \text{ (1 d.p.)} \end{align}

(ii)

\begin{align} \text{When } p & = 7.4 \text{ and } q = 0.2, \\ y & = (7.4)(0.2)^x \\ & = 7.4(0.2)^x \end{align}

(i)

\begin{align} Y & = mX + c \\ \\ m & = {8 - 2 \over 1 - 4} \\ & = -2 \\ \\ \therefore Y & = -2X + c \\ \\ \\ \text{When } & X = 1 \text{ and } Y = 8, \\ 8 & = -2(1) + c \\ 8 & = -2 + c \\ 8 + 2 & = c \\ \\ \therefore \text{Linear form: } \phantom{0} Y & = -2X + 10 \\ \\ \\ \text{Since } Y = y\sqrt{x} & \text{ and } X = x, \\ \\ y\sqrt{x} & = -2x + 10 \\ {1 \over \sqrt{x}}(y\sqrt{x}) & = {1 \over \sqrt{x}}(-2x + 10) \\ y & = {-2x \over \sqrt{x}} + {10 \over \sqrt{x}} \\ y & = -2\sqrt{x} + {10 \over \sqrt{x}} \end{align}

(ii)

\begin{align} &\text{When } x = 16, \\ y & = -2\sqrt{16} + {10 \over \sqrt{16}} \\ & = -{11 \over 2} \end{align}

(iii)

Since $X = x$, at point $C$, $x = 9$
\begin{align} & \text{When } x = 9, \\ y & = -2\sqrt{9} + {10 \over \sqrt{9}} \\ & = -6 + {10 \over 3} \\ & = -{8 \over 3} \end{align}

(i)

\begin{align} m & = {3 - (-3) \over 1 - (-1)} \\ & = 3 \\ \\ \therefore Y & = 3X + c \\ \\ \text{When } & X = 1 \text{ and } Y = 3, \\ 3 & = 3(1) + c \\ 3 & = 3 + c \\ 3 - 3 & = c \\ 0 & = c \\ \\ \text{Linear equation: } \phantom{0} Y & = 3X \\ \\ \\ \text{Since } Y = y - x & \text{ and } X = x^3, \\ \\ y - x & = 3x^3 \\ \\ y & = 3x^3 + x \end{align}

(ii)

\begin{align} & \text{When } x = 2, \\ y & = 3(2)^3 + (2) \\ & = 26 \end{align}

(iii)

\begin{align} m + 2n & = 56 \\ m & = 56 - 2n \phantom{000} \text{ --- (1)} \\ \\ n^2 - m & = 568 \phantom{000} \text{ --- (2)} \\ \\ \\ \text{Substitute } & \text{(1) into (2),} \\ n^2 - (56 - 2n) & = 568 \\ n^2 - 56 + 2n - 568 & = 0 \\ n^2 + 2n - 624 & = 0 \\ (n + 26)(n - 24) & = 0 \end{align} \begin{align} n + 26 & = 0 \phantom{-00} &\text{or} \phantom{0000} n - 24 & = 0 \\ n & = -26 \phantom{0} & \phantom{or0000-24} n & = 24 \\ \\ \text{Substitute } & \text{into (1),} & \text{Substitute } & \text{into (1),} \\ \\ m & = 56 - 2(-26) & m & = 56 - 2(24) \\ & = 108 & & = 8 \\ \end{align} $$ \therefore C(108, -26) \text{ [Reject]} \text{ or } C(8, 24) $$
The coordinate (108, -26) is rejected as the line is an upward sloping line and the Y-coordinate must be positive when X = 108
\begin{align} \text{Since } Y = y - x & \text{ and } X = x^3, \\ \\ x^3 & = 8 \\ x & = \sqrt[3]{8} \\ & = 2 \\ \\ y - x & = 24 \\ y - (2) & = 24 \\ y & = 24 + 2 \\ & = 26 \end{align}

(i)

\begin{align} m & = {28 - 10 \over 9 - 3} \\ & = 3 \\ \\ \therefore Y & = 3X + c \\ \\ \\ \text{Substitute } & X = 3 \text{ and } Y = 10, \\ 10 & = 3(3) + c \\ 10 & = 9 + c \\ 10 - 9 & = c \\ 1 & = c \\ \\ \text{Linear equation: } \phantom{0} Y & = 3X + 1 \\ \\ \text{Since } Y = {y \over x^2} & \text{ and } X = {1 \over x}$, \\ \\ {y \over x^2} & = 3 \left(1 \over x\right) + 1 \\ {y \over x^2} & = {3 \over x} + 1 \\ y & = x^2 \left({3 \over x} + 1\right) \\ y & = 3x + x^2 \end{align}

(ii)

\begin{align} \text{When } & x = {1 \over \sqrt{2}}, \\ y & = 3\left(1 \over \sqrt{2}\right) + \left(1 \over \sqrt{2}\right)^2 \\ & = {3 \over \sqrt{2}} + {1 \over 2} \\ & = {3 \over \sqrt{2}} \times {\sqrt{2} \over \sqrt{2}} + {1 \over 2} \phantom{000000} [\text{Rationalise denominator}] \\ & = {3\sqrt{2} \over 2} + {1 \over 2} \\ & = {3\sqrt{2} + 1 \over 2} \end{align}

(a)

\begin{align} x & = {a \over y - b} \\ x(y - b) & = a \\ xy - bx & = a \\ xy & = bx + a \\ {1 \over x} (xy) & = {1 \over x}(bx + a) \\ y & = b + {a \over x} \\ y & = a\left(1 \over x\right) + b \\ \\ \therefore Y & = aX + b \\ \\ \text{where } Y & = y \\ X & = {1 \over x} \\ m & = a \\ c & = b \end{align}
(Other possible answer: $Y = xy$, $X = x$, $m = b$ and $c = a$)

(b)

\begin{align} y & = ab^{1- x} \\ \\ \text{Take } \lg & \text{ of both sides,} \\ \lg y & = \lg (ab^{1-x}) \\ \lg y & = \lg a + \lg b^{1- x} \phantom{000000/} [\text{Product law (logarithms)}] \\ \lg y & = \lg a + (1 - x)\lg b \phantom{000} [\text{Power law (logarithms)}] \\ \lg y & = \lg a + \lg b - x\lg b \\ \lg y & = (-\lg b)x + \lg a + \lg b \\ \lg y & = (-\lg b)x + \lg ab \phantom{0000} [\text{Product law (logarithms)}] \\ \\ \therefore Y & = (-\lg b)X + \lg ab \\ \\ \text{where } Y & = \lg y \\ X & = x \\ m & = -\lg b \\ c & = \lg ab \end{align}
(Other possible answer by using $\ln$ instead of $\lg$: $Y = \ln y$ and $X = x$)

(c)

\begin{align} ae^y & = b^{2x} \\ \\ \text{Take } \ln & \text{ of both sides,} \\ \ln (ae^y) & = \ln b^{2x} \\ \ln a + \ln e^y & = \ln b^{2x} \phantom{0000000000} [\text{Product law (logarithms)}] \\ \ln a + y\ln e & = 2x(\ln b) \phantom{00000000} [\text{Power law (logarithms)}] \\ \ln a + y(1) & = (2\ln b)x \\ \ln a + y & = (\ln b^2)x \phantom{00000000} [\text{Power law (logarithms)}] \\ y & = (\ln b^2)x - \ln a \\ \\ \therefore Y & = (\ln b^2)X - \ln a \\ \\ \text{where } Y & = y \\ X & = x \\ m & = \ln b^2 \\ c & = -\ln a \end{align}

\begin{align} y & = p(x + 1)^q \\ \\ \text{Take } \lg & \text{ of both sides,} \\ \lg y & = \lg [p(x + 1)^q] \\ \lg y & = \lg p + \lg (x + 1)^q \phantom{00000} [\text{Product law (logarithms)}] \\ \lg y & = \lg p + q\lg (x + 1) \phantom{0000/} [\text{Power law (logarithms)}] \\ \lg y &= q\lg(x + 1) + \lg p \\ \\ \therefore Y & = qX + \lg p \\ \\ \text{where } Y & = \lg y \\ X & = \lg (x + 1) \\ m & = q \\ c & = \lg p \\ \\ \therefore \text{Plot } & \lg y \text{ against } \lg (x + 1) \end{align}
(Other answer by using $\ln$ instead of $\lg$: Plot $\ln y$ against $\ln (x + 1)$ )

(i)

\begin{align} Y & = mX + c \\ \\ \text{When } m = 3 & \text{ and } c = 2, \\ Y & = 3X + 2 \\ \\ \\ \text{Since } Y = y & \text{ and } X = {y \over x}, \\ \\ y & = 3\left(y \over x\right) + 2 \\ y & = {3y \over x} + 2 \\ x(y) & = x\left({3y \over x} + 2\right) \\ xy & = 3y + 2x \\ xy - 3y & = 2x \\ y(x - 3) & = 2x \\ y & = {2x \over x - 3} \\ \\ \text{Comparing with } & \phantom{0} y = {ax \over x - b}, \\ \\ a & = 2, \phantom{.} b = 3 \end{align}

(ii)

\begin{align} y & = {2x \over x - 3} \\ \\ \text{When } & x = y, \\ x & = {2x \over x - 3} \\ x(x - 3) & = 2x \\ x^2 - 3x & = 2x \\ x^2 - 5x & = 0 \\ x(x - 5) & = 0 \\ \\ x = 0 \phantom{00} &\text{or} \phantom{00} x - 5 = 0 \\ & \phantom{or00-5} x = 5 \end{align}

(i)

\begin{align} m & = {3{1 \over 2} - 1 \over 8 - 3} \\ & = {1 \over 2} \\ \\ \therefore Y & = {1 \over 2}X + c \\ \\ \\ \text{When } X = 3 & \text{ and } Y = 1, \\ 1 & = {1 \over 2}(3) + c \\ 1 & = {3 \over 2} + c \\ 1 - {3 \over 2} & = c \\ -{1 \over 2} & = c \\ \\ \text{Linear equation: } \phantom{0} Y & = {1 \over 2}X - {1 \over 2} \\ \\ \text{Since } Y = xy & \text{ and } X = {1 \over x}, \\ \\ xy & = {1 \over 2}\left(1 \over x\right) - {1 \over 2} \\ xy & = {1 \over 2x} - {1 \over 2} \\ 2(xy) & = 2\left({1 \over 2x} - {1 \over 2}\right) \\ 2xy & = {1 \over x} - 1 \text{ (Shown)} \end{align}

(ii)

\begin{align} \text{When } & x = 2, \\ 2(2)y & = {1 \over 2} - 1 \\ 4y & = -{1 \over 2} \\ y & = {-{1 \over 2} \over 4} \\ & = -{1 \over 8} \end{align}

(iii)

\begin{align} 2xy & = {1 \over x} - 1 \phantom{00000} \text{ --- (1)} \\ \\ y & = x^2 \phantom{00000000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 2x(x^2) & = {1 \over x} - 1 \\ 2x^3 & = {1 \over x} - 1 \\ 2x^4 & = 1 - x \\ 2x^4 + x - 1 & = 0 \\ \\ \text{Let } f(x) & = 2x^4 + x - 1 \\ \\ f(-1) & = 2(-1)^4 + (-1) - 1 \\ & = 0 \\ \\ \text{By Factor } & \text{theorem, } \phantom{.} x + 1 \text{ is a factor} \\ \\ \therefore \text{Integer value of } x & = -1 \end{align}

\begin{align} y = x^7 & - 14x^5 + 49x^3 - 35x \text{ is not a linear function} \\ \\ \\ & \text{When } x = 4, \\ y & = (4)^7 - 14(4)^5 + 49(4)^3 - 35(4) \\ y & = 5044 \\ \\ \text{No, } & (4, 4) \text{ does not lie on the graph} \\ \\ \\ & \text{When } x = -4, \\ y & = (-4)^7 - 14(-4)^5 + 49(-4)^3 - 35(-4) \\ y & = -5044 \\ \\ \text{No, } & (-4, -4) \text{ does not lie on the graph} \end{align}