Functions and Graphs - Estimating Gradients from Tangents

Functions and Graphs - Estimating Gradients from Tangents

Secondary 3

The variables \(x\) and \(y\) are connected by the equation \(y = x^3 - 3x^2 - 1\).

The table below shows some corresponding values of \(x\) and \(y\).

\(x\)\(-2\)\(-1\)\(0\)\(1\)\(2\)\(3\)\(4\)
\(y\)\(-21\)\(-5\)\(-1\)\(-3\)\(a\)\(-1\)\(15\)
  1. Calculate the value of \(a\).
  2. Taking \(2\) cm to represent \(1\) unit on the \(x\)-axis and \(2\) cm to represent \(5\) units on the \(y\)-axis, draw the graph of \(y = x^3 - 3x^2 - 1\) for the range of \(-2 \leq x \leq 4\).
  3. Using your graph, find the value of \(x\) when \(y = 10\).
  4. Using your graph, solve the equation
    1. \(x^3 - 3x^2 - 2 = 0\),
    2. \(x^3 - 3x^2 - 2x + 2 = 0\).
  5. By drawing a tangent, find the gradient of the curve where \(x = 2.5\).

Solution:

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Answer:(a) \(-5\). (c) \(3.77\). (d)(i) \(3.20\); (ii) \(-1,0.59,3.41\). (e) About \(3.75\).

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