(a)

\begin{align} {dy \over dx} & = 3x^2 + 6x \\ \\ \text{When } & x = -1, \\ {dy \over dx} & = 3(-1)^2 + 6(-1) \\ & = -3 \\ \\ y & = mx + c \\ y & = -3x + c \\ \\ \text{Using } & (-1, 2), \\ 2 & = -3(-1) + c \\ 2 & = 3 + c \\ -1 & = c \\ \\ \text{Eqn of tangent:} & \phantom{0} y = -3x - 1 \end{align}

(b)

\begin{align} y & = x + {2 \over x} \\ & = x + 2x^{-1} \\ \\ {dy \over dx} & = 1 + (2)(-1)(x)^{-2} \\ & = 1 - 2 x^{-2} \\ & = 1 - {2 \over x^2} \\ \\ \text{When } & x = 1, \\ {dy \over dx} & = 1 - {2 \over (1)^2} \\ & = -1 \\ \\ y & = mx + c \\ y & = -x + c \\ \\ \text{Using } & (1, 3), \\ 3 & = -1 + c \\ 4 & = c \\ \\ \text{Eqn of tangent:} & \phantom{0} y = -x + 4 \end{align}

(a)

\begin{align} \text{Substitute } & x = -1 \text{ into eqn of curve,} \\ y & = (-1)^3 - 5(-1) + 8 \\ & = -1 + 5 + 8 \\ & = 12 \\ \\ \text{Coordinates } & \text{of point is (-1, 12)} \\ \\ {dy \over dx} & = 3x^2 - 5 \\ \\ \text{When } & x = -1, \\ {dy \over dx} & = 3(-1)^2 - 5 \\ & = -2 \\ \\ \text{Gradient of normal} \times -2 & = -1 \\ \text{Gradient of normal} & = {-1 \over -2} \\ & = {1 \over 2} \\ \\ y & = mx + c \\ y & = {1 \over 2}x + c \\ \\ \text{Using } & (-1, 12), \\ 12 & = {1 \over 2}(-1) + c \\ 12 & = -{1 \over 2} + c \\ {25 \over 2} & = c \\ \\ \text{Eqn of normal: } \phantom{0} y & = {1 \over 2}x + {25 \over 2} \\ 2y & = x + 25 \end{align}

(b)

\begin{align} \text{Substitute } & y = 10 \text{ into eqn of curve,} \\ 10 & = 2x + {8 \over x} \\ 10x & = 2x^2 + 8 \phantom{00000000} [\text{Multiply by } x] \\ 0 & = 2x^2 - 10x + 8 \\ 0 & = x^2 - 5x + 4 \\ 0 & = (x - 4)(x - 1) \\ \\ x - 4 = 0 \phantom{00} & \text{or} \phantom{000} x - 1 = 0 \\ x = 4 \phantom{00} & \phantom{or000-1} x = 1 \text{ (Reject, since } x > 1) \\ \\ \text{Coordinates } & \text{of point is (4, 10)} \\ \\ y & = 2x + 8x^{-1} \\ \\ {dy \over dx} & = 2 + 8(-1)x^{-2} \\ & = 2 - 8x^{-2} \\ & = 2 - {8 \over x^2} \\ \\ \text{When } & x = 4, \\ {dy \over dx} & = 2 - {8 \over (4)^2} \\ & = {3 \over 2} \\ \\ \text{Gradient of normal} \times {3 \over 2} & = -1 \\ \text{Gradient of normal} & = -1 \div {3 \over 2} \\ & = -{2 \over 3} \\ \\ y & = mx + c \\ y & = -{2 \over 3}x + c \\ \\ \text{Using } & (4, 10), \\ 10 & = -{2 \over 3}(4) + c \\ 10 & = -{8 \over 3} + c \\ {38 \over 3} & = c \\ \\ \text{Eqn of normal: } \phantom{0} y & = -{2 \over 3}x + {38 \over 3} \\ 3y & = -2x + 38 \end{align}

(a)

\begin{align} \text{Substitute } & x = 3 \text{ into eqn of curve,} \\ y & = {(3)^2 + 5 \over 3 + 1} \\ & = {7 \over 2} \\ \\ \text{Coordinates } & \text{of point is } \left(3, {7 \over 2}\right) \end{align} \begin{align} u & = x^2 + 5 &&& v & = x + 1 \\ {du \over dx} & = 2x &&& {dv \over dx} & = 1 \end{align} \begin{align} {dy \over dx} & = {(x + 1)(2x) - (x^2 + 5)(1) \over (x + 1)^2} \phantom{00000} [\text{Quotient rule}] \\ & = {2x(x + 1) - (x^2 + 5) \over (x + 1)^2} \\ & = {2x^2 + 2x - x^2 - 5 \over (x + 1)^2} \\ & = {x^2 + 2x - 5 \over (x + 1)^2} \\ \\ {dy \over dx} & = {(3)^2 + 2(3) - 5 \over [(3) + 1]^2} \\ & = {5 \over 8} \\ \\ y & = mx + c \\ y & = {5 \over 8}x + c \\ \\ \text{Using } & \left(3, {7 \over 2}\right), \\ {7 \over 2} & = {5 \over 8}(3) + c \\ {7 \over 2} & = {15 \over 8} + c \\ {13 \over 8} & = c \\ \\ \text{Eqn of tangent: } \phantom{0} y & = {5 \over 8}x + {13 \over 8} \\ 8y & = 5x + 13 \\ \\ \\ \text{Gradient of tangent} & = {5 \over 8} \\ \\ \text{Gradient of normal} & = {-1 \over {5 \over 8}} \\ & = -{8 \over 5} \\ \\ y & = mx + c \\ y & = -{8 \over 5}x + c \\ \\ \text{Using } & \left(3, {7 \over 2}\right), \\ {7 \over 2} & = -{8 \over 5}(3) + c \\ {7 \over 2} & = -{24 \over 5} + c \\ {83 \over 10} & = c \\ \\ \text{Eqn of normal: } \phantom{0} y & = -{8 \over 5}x + {83 \over 10} \\ 10y & = -16x + 83 \end{align}

(b)

\begin{align} \text{Substitute } & y = 3 \text{ into eqn of curve,} \\ 3 & = \sqrt{1 - 2x} \\ (3)^2 & = (\sqrt{1 - 2x})^2 \\ 9 & = 1 - 2x \\ 2x & = 1- 9 \\ 2x & = -8 \\ x & = {-8 \over 2} \\ x & = -4 \\ \\ \text{Coordinates } & \text{of point is } (-4, 3) \\ \\ y & = \sqrt{1 - 2x} \\ & = (1 - 2x)^{1 \over 2} \\ \\ {dy \over dx} & = {1 \over 2} (1 - 2x)^{-{1 \over 2}} . (-2) \phantom{00000} [\text{Chain rule}] \\ & = (-1)(1 - 2x)^{-{1 \over 2}} \\ & = -{1 \over \sqrt{1 - 2x}} \\ \\ \text{When } & x = -4, \\ {dy \over dx} & = - {1 \over \sqrt{1 - 2(-4)}} \\ & = -{1 \over 3} \\ \\ y & = mx + c \\ y & = -{1 \over 3}x + c \\ \\ \text{Using } & (-4, 3), \\ 3 & = -{1 \over 3}(-4) + c \\ 3 & = {4 \over 3} + c \\ {5 \over 3} & = c \\ \\ \text{Eqn of tangent: } \phantom{0} y & = -{1 \over 3}x + {5 \over 3} \\ 3y & = -x + 5 \\ \\ \\ \text{Gradient of tangent} & = -{1 \over 3} \\ \\ \text{Gradient of normal} & = {-1 \over -{1 \over 3}} \\ & = 3 \\ \\ y & = mx + c \\ y & = 3x + c \\ \\ \text{Using } & (-4, 3), \\ 3 & = 3(-4) + c \\ 3 & = -12 + c \\ 15 & = c \\ \\ \text{Eqn of normal: } \phantom{0} y & = 3x + 15 \end{align}

(i)

\begin{align} {dy \over dx} & = {2 - x \over \sqrt{4x - x^2 + 1} } \\ \\ \text{When } & x = 1, \\ {dy \over dx} & = {2 - (1) \over \sqrt{4(1) - (1)^2 + 1}} \\ & = {1 \over 2} \\ \\ y & = mx + c \\ y & = {1 \over 2}x + c \\ \\ \text{Using } & P(1, 2), \\ 2 & = {1 \over 2}(1) + c \\ 2 & = {1 \over 2} + c \\ {3 \over 2} & = c \\ \\ \text{Eqn of tangent: } \phantom{0} y & = {1 \over 2}x + {3 \over 2} \\ 2y & = x + 3 \end{align}

(ii)

\begin{align} \text{Gradient of tangent} & = {1 \over 2} \\ \\ \text{Gradient of normal} & = {-1 \over {1 \over 2}} \\ & = -2 \\ \\ y & = mx + c \\ y & = -2x + c \\ \\ \text{Using } & P(1, 2), \\ 2 & = -2(1) + c \\ 2 & = -2 + c \\ 4 & = c \\ \\ \text{Eqn of normal: } \phantom{0} y & = -2x + 4 \end{align}

The point where the curve crosses the x-axis is the same as the x-intercept of the curve, where y = 0.

\begin{align} \text{Substiute } & y = 0 \text{ into eqn of curve,} \\ 0 & = {3x + 1 \over 1 - x} \\ 0 & = 3x + 1 \\ -1 & = 3x \\ -{1 \over 3} & = x \\ \\ x \text{-intercept} & \text{ is } \left(-{1 \over 3}, 0 \right) \end{align}
\begin{align} u & = 3x + 1 &&& v & = 1 - x \\ {du \over dx} & = 3 &&& {dv \over dx} & = -1 \end{align} \begin{align} {dy \over dx} & = {(1 - x)(3) - (3x + 1)(-1) \over (1 - x)^2} \phantom{00000} [\text{Quotient rule}] \\ & = {3(1 - x) + (3x + 1) \over (1 - x)^2} \\ & = {3 - 3x + 3x + 1 \over (1 - x)^2} \\ & = {4 \over (1 - x)^2} \\ \\ \text{When } & x = -{1 \over 3}, \\ {dy \over dx} & = {4 \over \left[1 - \left(-{1 \over 3}\right)\right]^2} \\ & = {9 \over 4} \\ \\ \text{Gradient of normal} & = {-1 \over {9 \over 4}} \\ & = -{4 \over 9} \\ \\ y & = mx + c \\ y & = -{4 \over 9}x + c \\ \\ \text{Using } & \left(-{1 \over 3}, 0 \right), \\ 0 & = -{4 \over 9}\left(-{1 \over 3}\right) + c \\ 0 & = {4 \over 27} + c \\ -{4 \over 27} & = c \\ \\ \text{Eqn of normal:} \phantom{0} y & = -{4 \over 9}x - {4 \over 27} \\ 27y & = -12x - 4 \end{align}

Recall that for perpendicular lines, $m_1 \times m_2 = -1$

\begin{align} 4y + x & = 2 \\ 4y & = -x + 2 \\ y & = -{1 \over 4}x + {1 \over 2} \\ \\ \therefore \text{Gradient of line} & = -{1 \over 4} \\ \\ \text{Gradient of tangent} & = {-1 \over -{1 \over 4}} \\ & = 4 \\ \\ {dy \over dx} & = 3(2)x - 2 \\ & = 6x - 2 \\ \\ \text{When } & {dy \over dx} = 4, \\ 4 & = 6x - 2 \\ 4 + 2 & = 6x \\ 6 & = 6x \\ {6 \over 6} & = x \\ 1 & = x \\ \\ \text{Substitute } & x = 1 \text{ into eqn of curve,} \\ y & = 3(1)^2 - 2(1) + 5 \\ & = 6 \\ \\ \text{Coordinates } & \text{of point is (1, 6)} \\ \\ y & = mx + c \\ y & = 4x + c \\ \\ \text{Using } & (1, 6), \\ 6 & = 4(1) + c \\ 6 & = 4 + c \\ 2 & = c \\ \\ \text{Eqn of tangent:} & \phantom{0} y = 4x + 2 \end{align}

Recall that parallel lines have the same gradient.

\begin{align} \text{For } & y = x^3 + 2x^2 - 4x + 5, \\ {dy \over dx} & = 3(x)^2 + 2(2)x - 4 \\ & = 3x^2 + 4x - 4 \\ \\ \text{For } & y = {2 \over 3}x^3 + 3x^2 - x - 1, \\ {dy \over dx} & = {2 \over 3}(3)(x)^2 + 3(2)(x) - 1 \\ & = 2x^2 + 6x - 1 \\ \\ \text{Since tan} & \text{gents are parallel,} \\ 3x^2 + 4x - 4 & = 2x^2 + 6x - 1 \\ 3x^2 - 2x^2 + 4x - 6x - 4 + 1 & = 0 \\ x^2 - 2x - 3 & = 0 \\ (x - 3)(x + 1) & = 0 \\ \\ x - 3 = 0 \phantom{000} & \text{or} \phantom{000} x + 1 = 0 \\ x = 3 \phantom{000} & \phantom{or000+1} x = - 1 \\ \\ \\ \therefore \text{Eqn of vertical lines:} & \phantom{0} x = 3 \text{ and } x = -1 \end{align}

(i)

\begin{align} {dy \over dx} & = a(5)(x)^4 \\ & = 5ax^4 \\ \\ \text{When } x = 1 & \text{ and } {dy \over dx} = {3 \over 2}, \\ {3 \over 2} & = 5a(1)^4 \\ {3 \over 2} & = 5a \\ {3 \over 2} \div 5 & = a \\ {3 \over 10} & = a \\ \\ y & = {3 \over 10}x^5 \\ \\ \text{Using } & (1, b), \\ b & = {3 \over 10}(1)^5 \\ b & = {3 \over 10} (1) \\ b & = {3 \over 10} \\ \\ \therefore a & = {3 \over 10}, b = {3 \over 10} \end{align}

(ii) Note origin refers to O(0, 0)

\begin{align} {dy \over dx} & = 5ax^4 \\ & = 5\left(3 \over 10\right)x^4 \\ & = {3 \over 2}x^4 \\ \\ \text{When } & x = 0, \\ {dy \over dx} & = {3 \over 2}(0)^4 \\ & = 0 \\ \\ \implies \text{Tangent is a} & \text{ horizontal line through } O(0, 0) \\ \\ \text{Eqn of tangent:} & \phantom{0} y = 0 \end{align}

(a)

\begin{align} y & = 3x + k \\ \\ \text{Comparing with } & y = mx + c, m = 3 \\ \\ \text{Gradient of tangent} & = 3 \\ \\ {dy \over dx} & = 2(2)x - 5 \\ & = 4x - 5 \\ \\ \text{When } & {dy \over dx} = 3, \\ 3 & = 4x - 5 \\ 3 + 5 & = 4x \\ 8 & = 4x \\ {8 \over 4} & = x \\ 2 & = x \\ \\ \text{Substitute } & x = 2 \text{ into eqn of curve,} \\ y & = 2(2)^2 - 5(2) + 3 \\ & = 1 \\ \\ \text{Coordinates} & \text{ of point is (2, 1)} \\ \\ \text{Eqn of tangent: } & y = 3x + k \\ \\ \text{Using } & (2, 1), \\ 1 & = 3(2) + k \\ 1 & = 6 + k \\ -5 & = k \end{align}

(b)

\begin{align} 2y + x & = l \\ 2y & = -x + 1 \\ y & = -{1 \over 2}x + {1 \over 2}l \\ \\ \text{Comparing with } & y = mx + c, m = -{1 \over 2} \\ \\ \text{Gradient of normal} & = -{1 \over 2} \\ \\ \text{Gradient of tangent} & = {-1 \over -{1 \over 2}} \\ & = 2 \\ \\ {dy \over dx} & = (2)(x) - 2 \\ & = 2x - 2 \\ \\ \text{When } & {dy \over dx} = 2, \\ 2 & = 2x - 2 \\ 2 + 2 & = 2x \\ 4 & = 2x \\ {4 \over 2} & = x \\ 2 & = x \\ \\ \text{Substitute } & x = 2 \text{ into eqn of curve,} \\ y & = (2)^2 -2(2) + 3 \\ & = 3 \\ \\ \text{Coordinates} & \text{ of point is (2, 3)} \\ \\ \text{Eqn of normal: } & 2y + x = l \\ \\ \text{Using } & (2, 3), \\ 2(3) + 2 & = l \\ 8 & = l \end{align}

(i)

\begin{align} {dy \over dx} & = p(3)(x)^2 \\ & = 3px^2 \\ \\ \text{When } & x = 2, \\ {dy \over dx} & = 3p(2)^2 \\ & = 3p(4) \\ & = 12p \end{align}

(ii)

\begin{align} \require{cancel} \text{Substitute } & x = 2 \text{ into eqn of curve,} \\ y & = p(2)^3 \\ & = p(8) \\ & = 8p \\ \\ \implies & A(2, 8p) \\ \\ \text{Gradient of AB } & = {y_2 - y_1 \over x_2 - x_1} \\ & = {8p - 8 \over 2 - 0} \\ & = {8p - 8 \over 2} \\ & = {\cancel{2}(4p - 4) \over \cancel{2}} \\ & = 4p - 4 \end{align}

(iii) Since line segment AB is the tangent to the curve at A, the gradient of the curve at A is equal to the gradient of line segment AB.

In other words, the answer for (i) is equal to the answer for (ii).

\begin{align} 12p & = 4p - 4 \\ 12p - 4p & = -4 \\ 8p & = -4 \\ p & = {-4 \over 8} \\ p & = -{1 \over 2} \end{align}

(i)

\begin{align} {dy \over dx} & = 2(2)(x) - k \\ & = 4x - k \\ \\ \text{When } & x = 1, \\ {dy \over dx} & = 4(1) - k \\ & = 4 - k \end{align}

(ii)

\begin{align} \text{Substitute } & x = 1 \text{ into eqn of curve,} \\ y & = 2(1)^2 - k(1) + 3 \\ & = 5 - k \\ \\ \implies & A(1, 5 - k) \\ \\ \text{Gradient of tangent } AB & = {y_2 - y_1 \over x_2 - x_1} \\ & = {(5 - k) - 1 \over 1 - 5} \\ & = {5 - k - 1 \over -4} \\ & = {4 - k \over -4} \\ \\ \text{From (i), gradient of} & \text{ tangent } AB = 4 - k, \\ 4 - k & = {4 - k \over -4} \\ -4(4 - k) & = 4 - k \\ -16 + 4k & = 4 - k \\ 4k + k & = 4 + 16 \\ 5k & = 20 \\ k & = {20 \over 5} \\ k & = 4 \end{align}

(i)

\begin{align} \text{Substitute } & x = a \text{ into eqn of curve,} \\ y & = a + (a)^2 \\ & = a + a^2 \\ \\ \text{Coordinates} & \text{ point is } (a, a + a^2) \\ \\ {dy \over dx} & = 1 + (2)(x) \\ & = 1 + 2x \\ \\ \text{When } & x = a, \\ {dy \over dx} & = 1 + 2(a) \\ & = 1 + 2a \\ \\ y & = mx + c \\ y & = (1 + 2a)x + c \\ \\ \text{Using } & (a, a + a^2), \\ a + a^2 & = (1 + 2a)(a) + c \\ a + a^2 & = a + 2a^2 + c \\ a - a + a^2 - 2a^2 & = c \\ -a^2 & = c \\ \\ \text{Eqn of tangent:} & \phantom{0} y = (1 + 2a)x - a^2 \end{align}

(ii)

\begin{align} \text{Eqn of tangent:} & \phantom{0} y = (1 + 2a)x - a^2 \\ \\ \text{Using } & P(2, -3), \\ -3 & = (1 + 2a)(2) - a^2 \\ -3 & = 2 + 4a - a^2 \\ a^2 -4a - 3 - 2 & = 0 \\ a^2 - 4a - 5 & = 0 \\ (a - 5)(a + 1) & = 0 \\ \\ a - 5 = 0 \phantom{000} & \text{or} \phantom{000} a + 1 = 0 \\ a = 5 \phantom{000} & \phantom{or000+1} a = - 1 \end{align}

Hence

\begin{align} \text{For } & a = 5, &&& \text{For } & a = -1, \\ y & = [1 + 2(5)]x - (5)^2 &&& y & = [1 + 2(-1)]x - (-1)^2 \\ y & = 11x - 25 &&& y & = -x - 1 \end{align}

(i)

\begin{align} \text{Substitute } & x = 1 \text{ into eqn of curve,} \\ y & = {4 \over a^2} + 1 \\ \\ \text{Coordinates} & \text{ of point is } \left(a, {4 \over a^2} + 1\right) \\ \\ {dy \over dx} & = {d \over dx}[4(x)^{-2} + 1] \\ & = 4(-2)(x)^{-3} \\ & = -{8 \over x^3} \\ \\ \text{When } & x = a, \\ {dy \over dx} & = -{8 \over a^3} \\ \\ y & = mx + c \\ y & = -{8 \over a^3} x + c \\ \\ \text{Using } & \left(a, {4 \over a^2} + 1\right), \\ {4 \over a^2} + 1 & = -{8 \over a^3} (a) + c \\ {4 \over a^2} + 1 & = -{8 \over a^2} + c \\ {4 \over a^2} + {8 \over a^2} + 1 & = c \\ {12 \over a^2} + 1 & = c \\ \\ \text{Eqn of tangent:} \phantom{0} y & = -{8 \over a^3} x + {12 \over a^2} + 1 \\ a^2 (y) & = a^2 \left( -{8 \over a^3} x + {12 \over a^2} + 1 \right) \\ a^2 y & = -{8 \over a}x + 12 + a^2 \end{align}

(ii)

\begin{align} \text{Gradient of } PQ & = {y_2 - y_1 \over x_2 - x_1} \\ & = {0 - b \over b - 0} \\ & = -1 \\ \\ \text{Gradient of tangent} & = \text{Gradient of } PQ \\ -{8 \over a^3} & = -1 \\ {8 \over a^3} & = 1 \\ 8 & = a^3 \\ \sqrt[3]{8} & = a \\ 2 & = a \\ \\ \text{Substitute } & a = 2 \text{ into eqn of tangent,} \\ y & = -{8 \over (2)^3} x + {12 \over (2)^2} + 1 \\ y & = (-1)x + 3 + 1 \\ y & = -x + 4 \\ \\ \text{Using } & P(b, 0), \\ 0 & = -b + 4 \\ b & = 4 \\ \\ \therefore a & = 2, b = 4 \end{align}

(i)

\begin{align} \text{Gradient of } BD & = {y_2 - y_1 \over x_2 - x_1} \\ & = {11 - 0 \over 0 - (- 2.2)} \\ & = 5 \\ \\ {dy \over dx} & = 2(x) + 7 \\ & = 2x + 7 \\ \\ \text{When } & {dy \over dx} = 5, \\ 5 & = 2x + 7 \\ 5 - 7 & = 2x \\ -2 & = 2x \\ {-2 \over 2} & = x \\ -1 & = x \\ \\ \text{Substitute } & x = -1 \text{ into eqn of curve,} \\ y & = (-1)^2 + 7(-1) + 12 \\ & = 6 \\ \\ \therefore & \phantom{.} A(-1, 6) \end{align}

(ii)

\begin{align} \text{Gradient of normal} & = {-1 \over \text{Gradient of tangent}} \\ & = {-1 \over 5} \\ & = -{1 \over 5} \\ \\ y & = mx + c \\ y & = -{1 \over 5}x + c \\ \\ \text{Using } & A(-1, 6), \\ 6 & = -{1 \over 5}(-1) + c \\ 6 & = {1 \over 5} + c \\ {29 \over 5} & = c \\ \\ \text{Eqn of normal: } & \phantom{0} y = -{1 \over 5}x + {29 \over 5} \\ \\ \text{When } & y = 0, \phantom{00000000} [C \text{ is the } x \text{-intercept}] \\ 0 & = -{1 \over 5}x + {29 \over 5} \\ {1 \over 5}x & = {29 \over 5} \\ x & = 29 \\ \\ \therefore & \phantom{.} C(29, 0) \end{align}

(iii) If BC is the base of the triangle, the dotted line is the height of the triangle.

Alternatively, you can use 'shoelace' method to find the area as well.

\begin{align} BC & = 29 - (-2.2) \\ & = 31.2 \text{ units} \\ \\ \text{Area of triangle} ABC & = {1 \over 2}\times\text{Base}\times\text{Height} \\ & = {1 \over} \times BC \times h \\ & = {1 \over 2}(31.2)(6) \\ & = 93.6 \text{ units}^2 \end{align}

\begin{align} x - 2y & = 4 \\ -2y & = -x + 4 \\ 2y & = x - 4 \\ y & = {1 \over 2}x - 2 \\ \\ \text{Gradient of line} & = {1 \over 2} \\ \\ \text{Gradient of tangent} & = {1 \over 2} \phantom{00000} [\text{Parallel lines}] \\ \\ (y - 2)^2 & = x \\ y - 2 & = \pm\sqrt{x} \\ y & = \pm\sqrt{x} + 2 \\ & = \pm (x)^{1 \over 2} + 2 \\ \\ {dy \over dx} & = \pm\left({1 \over 2}\right)(x)^{-{1 \over 2}} \\ & = \pm{1 \over 2}\left({1 \over \sqrt{x}}\right) \\ & = \pm {1 \over 2\sqrt{x}} \\ \\ \text{Since gradient of tan} & \text{gent} = {1 \over 2}, \text{ reject } {dy \over dx} = -{1 \over 2\sqrt{x}} \text{ and } y = -\sqrt{x} + 2 \\ \\ \text{When } & {dy \over dx} = {1 \over 2}, \\ {1 \over 2} & = {1 \over 2\sqrt{x}} \\ 2\sqrt{x} & = 2 \\ \sqrt{x} & = 1 \\ (\sqrt{x})^2 & = 1^2 \\ x & = 1 \\ \\ \text{Substitute } & x = 1 \text{ into } y = \sqrt{x} + 2, \\ y & = \sqrt{1} + 2 \\ & = 3 \\ \\ \text{Coordinates} & \text{ of point is (1, 3)} \\ \\ y & = mx + c \\ y & = {1 \over 2}x + c \\ \\ \text{Using } & (1, 3), \\ 3 & = {1 \over 2}(1) + c \\ 3 & = {1 \over 2} + c \\ {5 \over 2} & = c \\ \\ \text{Eqn of tangent: } \phantom{0} y & = {1 \over 2}x + {5 \over 2} \\ 2y & = x + 5 \\ 0 & = x - 2y + 5 \end{align}

\begin{align} x^3 + xy^2 & = 5 \\ xy^2 & = 5 - x^3 \\ y^2 & = {5 \over x} - x^2 \\ y & = \pm\sqrt{{5 \over x} - x^2} \\ \\ \text{Since } y \text{-coor} & \text{dinate of (1, 2) is 2,} \\ y & = \sqrt{{5 \over x} - x^2} \\ & = \left(5x^{-1} - x^2 \right)^{-1} \\ \\ {dy \over dx} & = {1 \over 2}[5(x)^{-1} - x^2]^{-{1 \over 2}}{d \over dx}[5(-1)(x)^{-2} - 2x] \phantom{00000000} [\text{Chain rule}]\\ & = {1 \over 2}[5(x)^{-1} - x^2]^{-{1 \over 2}}[-5x^{-2} - 2x] \\ & = {1 \over 2} \left[ 1 \over \sqrt{5(x)^{-1} - x^2} \right] [5(-1)(x)^{-2} - 2x] \\ & = {1 \over 2}\left[{-5(x)^{-2} - 2x \over \sqrt{5(x)^{-1} - x^2}}\right] \\ \\ \text{When } & x = 1, \\ {dy \over dx} & = {1 \over 2}\left[ {-5(1)^{-2} - 2(1) \over \sqrt{5(1)^{-1} - (1)^2}} \right] \\ & = {1 \over 2}\left({-7 \over 2}\right) \\ & = -{7 \over 4} \\ \\ \therefore \text{Gradient of tangent} & = -{7 \over 4} \end{align}

(i)

\begin{align} \text{Substitute } & x = 1 \text{ into eqn of curve,} \\ y & = (1)^3 + 5 \\ & = 6 \\ \\ \text{Coordinates} & \text{ of point is (1, 6)} \\ \\ {dy \over dx} & = 3(x)^2 \\ & = 3x^2 \\ \\ \text{When } & x = 1, \\ {dy \over dx} & = 3(1)^2 \\ & = 3 \\ \\ y & = mx + c \\ y & = 3x + c \\ \\ \text{Using } & (1, 6), \\ 6 & = 3(1) + c \\ 6 & = 3 + c \\ 3 & = c \\ \\ \text{Eqn of tangent:} & \phantom{0} y = 3x + 3 \end{align}

(ii)

\begin{align} \text{Eqn of tangent: } \phantom{0} y & = 3x + 3 \phantom{00} \text{--- (1)} \\ \text{Eqn of curve: } \phantom{0} y & = x^3 + 5 \phantom{00} \text{--- (2)} \\ \\\text{Substitute } & \text{(1) into (2),} \\ 3x + 3 & = x^3 + 5 \\ x^3 - 3x + 5 - 3 & = 0 \\ x^3 - 3x + 2 & = 0 \phantom{0000000} [\text{Cubic equation}] \\ \\ \text{Let } & f(x) = x^3 - 3x + 2 \\ \\ f(1) & = (1)^3 - 3(1) + 2 \\ & = 0 \\ \\ \text{By Factor theorem,} & \phantom{0} x - 1 \text{ is a factor} \end{align}
$$ \require{enclose} \begin{array}{rll} x^2 + x - 2\phantom{0000000}\\ x - 1 \enclose{longdiv}{x^3 + 0x^2 - 3x + 2\phantom{0}}\kern-.2ex \\ -\underline{(x^3 - x^2){\phantom{00000000}}} \\ x^2 - 3x + 2\phantom{0} \\ -\underline{(x^2 - x){\phantom{0000}}} \\ - 2x + 2\phantom{0} \\ -\underline{(-2x + 2)} \\ 0\phantom{0} \end{array} $$
\begin{align} \text{Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder} \\ x^3 - 3x + 2 & = (x - 1)(x^2 + x - 2) + 0 \\ & = (x - 1)(x - 1)(x + 2) \\ & = (x - 1)^2 (x + 2) \\ \\ x^3 - 3x + 2 & = 0 \\ (x - 1)^2(x + 2) & = 0 \\ \\ x - 1 = 0 \phantom{00000000(.} & \text{or} \phantom{000} x + 2 = 0 \\ x = 1 \text{ (Repeated)} & \phantom{or000+2} x = - 2 \\ \\ \\ \text{Substitute } & x = -2 \text{ into eqn of curve,} \\ y & = (-2)^3 + 5 \\ & = -3 \\ \\ \text{Coordinates} & \text{ is } (-2, -3) \end{align}