\begin{align*} y & = a \sqrt{x} + {b \over \sqrt{x} } \\ \sqrt{x} (y) & = \sqrt{x} \left( a \sqrt{x} + {b \over \sqrt{x} } \right) \\ \sqrt{x} y & = a x + b \\ \\ Y & = \sqrt{x} y \\ m & = a \\ X & = x \\ c & = b \end{align*}

\begin{align*} {y \over 1} & = {a \over x - b} \\ y(x - b) & = a \phantom{0000000000} [\text{Cross-multiply}] \\ xy - by & = a \\ xy & = by + a \\ \\ Y & = xy \\ m & = b \\ X & = y \\ c & = a \end{align*}

\begin{align*} y^b & = 10^{x + a} \\ \lg y^b & = \lg 10^{x + a} \\ b \lg y & = (x + a) \lg 10 \phantom{000000} [\text{Power law}] \\ b \lg y & = (x + a) (1) \\ b \lg y & = x + a \\ \lg y & = {1 \over b} (x + a) \\ \lg y & = {1 \over b}x + {a \over b} \\ \\ Y & = \lg y \\ m & = {1 \over b} \\ X & = x \\ c & = {a \over b} \end{align*}

\begin{align*} y & = ax^b + 5 \\ y - 5 & = ax^b \\ \ln (y - 5) & = \ln (ax^b) \phantom{0000000000} [\text{Can use } \lg \text{ as well}] \\ \ln (y - 5) & = \ln a + \ln x^b \phantom{0000000} [\text{Product law}] \\ \ln (y - 5) & = \ln a + b \ln x \phantom{000000.} [\text{Power law}] \\ \ln (y - 5) & = b \ln x + \ln a \\ \\ Y & = \ln (y - 5) \\ m & = b \\ X & = \ln x \\ c & = \ln a \end{align*}

\begin{align*} {y \over 1} & = {x \over ax + b} \\ y(ax + b) & = x \phantom{000000000000} [\text{Cross-multiply}] \\ ax + b & = {x \over y} \\ {1 \over x}(ax + b) & = {1 \over y} \\ a + b\left(1 \over x\right) & = {1 \over y} \\ \\ {1 \over y} & = b \left(1 \over x\right) + a \phantom{0000} [Y = mX + c] \\ \\ \text{Gradient of line} & = { 7 - 1 \over 2 - 5 } \\ \text{Gradient of line} & = -2 \\ \implies b & = -2 \\ \\ {1 \over y} & = -2 \left(1 \over x\right) + a \\ \\ \text{Using } & (5, 1), \\ 1 & = -2 (5) + a \\ 1 & = -10 + a \\ 11 & = a \\ \\ \therefore a & = 11, b = -2 \end{align*}

\begin{align*} y & = 1 + e^a x^b \\ y - 1 & = e^a x^b \\ \ln (y - 1) & = \ln (e^a x^b) \\ \ln (y - 1) & = \ln e^a + \ln x^b \phantom{000000} [\text{Product law}] \\ \ln (y - 1) & = a \ln e + b \ln x \phantom{00000} [\text{Power law}] \\ \ln (y - 1) & = a(1) + b \ln x \\ \ln (y - 1) & = b \ln x + a \phantom{00000000} [Y = mX + c] \\ \\ \text{Gradient of line} & = {5 - 1 \over 2 - 1} \\ & = 4 \\ \implies b & = 4 \\ \\ \ln (y - 1) & = 4 \ln x + a \\ \\ \text{Using } & (1, 1), \\ 1 & = 4(1) + a \\ 1 & = 4 + a \\ -3 & = a \\ \\ \therefore a & = -3, b = 4 \end{align*}

\begin{align} \text{Correct point: } & \left(6, 7 \right) \\ \\ {y \over x^2} & = 7 \\ {y \over 6^2} & = 7 \\ y & = 7(6^2) \\ & = 252 \end{align}

\begin{align} y & = ax^3 + bx^2 \\ y & = x^2(ax + b) \\ \text{Linear eqn: } \phantom{000} {y \over x^2} & = ax + b \\ \\ \text{where } Y & = {y \over x^2} \\ X & = x \\ m & = a \\ c & = b \\ \\ \text{From graph, vertical intercept, } c & = -4.9 \\ \therefore b & = -4.9 \\ \\ \text{Gradient, } m & = {7 - (-2.5) \over 6 - 1.2} \\ & = 1.97916 \\ & \approx 1.98 \\ \therefore a & \approx 1.98 \end{align}

\begin{align*} \text{From graph, when } x & = 2, {y \over x^2} = -0.9 \\ \\ {y \over (2)^2} & = -0.9 \\ {y \over 4} & = {-0.9 \over 1} \\ y & = 4(-0.9) \\ y & = -3.6 \end{align*}