Find \(k\), giving your answer correct to 4 decimal places.
Find the gradient of \(AE\), giving your answer correct to 4 decimal places.
The gradients of \(BE, CE\) and \(DE\), rounded to 4 decimal places, are 1.9748, 1.9975 and 1.9997 respectively.
State, giving a reason for your answer, what the values of the four gradients suggest about the gradient of the curve at the point \(E\).
Solution
(a) To find \(k\), substitute \(x = 2\) into the curve equation:
\(y = (2 - 3)\sqrt{2 + 1} + 3\)
\(y = (-1)\sqrt{3} + 3\)
\(y = 3 - \sqrt{3}\)
\(k = 1.2679\)
(b) The gradient of \(AE\) is given by:
\(\text{Gradient} = \frac{y_2 - y_1}{x_2 - x_1}\)
\(\text{Gradient} = \frac{3 - 1.2679}{3 - 2}\)
\(\text{Gradient} = 1.7321\)
(c) The gradients of \(BE, CE, DE\) are approaching 2 as \(x\) approaches 3. This suggests that the gradient of the curve at \(E\) is the limit of these gradients, which is approximately 2.