(–7) = 2(x 1) y – – + 7 = 2x 2 y – = 2x 9 y – 3. Calculate the point on = 3x2 4x 6 where the gradient is 2. y – – Solution: Step 1: Determine ; i.e: = 6x 4 – Step 2: You are the value of = 2 m i.e. 6x 4 = 2 – 6x = 6 = 1 x Step 3: Use the obtained in Step 2 above to obtain the corresponding x –value i.e: = 3(1)2 4(1) 6 y – – = 7 y – Step 4: down the coordinates ; i.e (1; 7) – Activity 7 1. Determine, the gradient of the curve at the given value of x Function of x a) = 2x = 3 f(x) x2 x – b) = 2x2 + 1 = 2 g(x) x3 x x – – c) = 3 = 1 xy x d) = (3x = 0 y x –1)2 e) = = 2 y x f) = = 1 y x 2. Determine the equation of the tangents to the curve at the given value of x a) = 2x2 + 4 , at = 2 b) = 4 at = 2 y x3 x xy x – – c) = 2x 8, at = 1 d) = at = 1 y x2 x y x3 x – – e) = 3x2 + 2x + 1 , at = 1 f)** = at = 2 y x y x – – 3. Determine the coordinates of the point P at which the gradient of the curve = is y x2 6. 4. Determine the equation of the tangent to the curve = 3x which is parallel to the y x2 – line 4 = 0. y x – – 5** Q(x ; is any point on the curve defined by = 5x. y) y x2 – a) Determine the equation of the tangent at Q which is parallel to the straight- line = 2x + 1 y b) Draw a sketch to illustrate the given information c) Calculate the co-ordinates of the point Q. d) Hence, determine the equation of the straight-line PQR
y – (–7) = 2(x – 1) y + 7 = 2x – 2 y = 2x – 9 3. Calculate the point on y = 3x2 – 4x – 6 where the gradient is 2. Solution: Step 1: Determine ; i.e: = 6x – 4 Step 2: You are the value of m = 2 i.e. 6x – 4 = 2 6x = 6 x = 1 Step 3: Use the x –value obtained in Step 2 above to obtain the corresponding i.e: y = 3(1)2 – 4(1) – 6 y = – 7 Step 4: down the coordinates ; i.e (1; – 7) Activity 7 1. Determine, the gradient of the curve at the given value of x Function of x a) f(x) = x2 – 2x x = 3 b) g(x) = x3 – 2x2 + x – 1 x = 2 c) xy = 3 x = 1 d) y = (3x –1)2 x = 0 e) y = x = 2 f) y = x = 1 2. Determine the equation of the tangents to the curve at the given value of x a) y = x3 – 2x2 + 4 , at x = 2 b) xy = 4 at x = – 2 c) y = x2 – 2x – 8, at x = 1 d) y = x3 at x = 1 e) y = – 3x2 + 2x + 1 , at x = – 1 f)** y = at x = 2 3. Determine the coordinates of the point P at which the gradient of the curve y = x2 is 6. 4. Determine the equation of the tangent to the curve y = 3x – x2 which is parallel to the line y – x – 4 = 0. 5** Q(x ; y) is any point on the curve defined by y = x2 – 5x. a) Determine the equation of the tangent at Q which is parallel to the straight- line y = 2x + 1 b) Draw a sketch to illustrate the given information c) Calculate the co-ordinates of the point Q. d) Hence, determine the equation of the straight-line PQR
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