Curve sketching
In order to sketch the curve of a function, you need to:
- Differentiate the function
- Set \(\frac{{dy}}{{dx}} = 0\)
- To put an expression into brackets. For example, 18x + 12y = 6(3x + 2y). Factorising is the reverse process to expanding. and then solve to find the \(x\)-coordinates of stationary points
- Find the nature of the curve using the \(x\)-coordinates in a Nature table to determine the shape of the curve to the left and right of the stationary points
- Substitute the \(x\)-coordinates of stationary points into the original equation of the curve to find the \(y\)-coordinates at those points
- Find where the curve cuts the \(y\)-axis by substituting \(x = 0\) into the original equation of the curve
- Find where the curve cuts the \(x\)-axis by substituting \(y = 0\) into the original equation of the curve
- Sketch the curve
Example
Sketch the curve \(y = {x^2} + 4x - 5\)
Solution
\(\frac{{dy}}{{dx}} = 2x + 4\)
Stationary Points occur when \(\frac{{dy}}{{dx}} = 0\)
\(2x + 4 = 0\)
\(2(x + 2) = 0\)
\(x + 2 = 0\)
\(x = - 2\)
\(\frac{{dy}}{{dx}} = 2( - 3) + 4 = - 6 + 4 = - 2\) (negative)
\(\frac{{dy}}{{dx}} = 2( - 2) + 4 = - 4 + 4 = 0\) (stationary)
\(\frac{{dy}}{{dx}} = 2( - 1) + 4 = - 2 + 4 = 2\) (positive)
y-coordinates
When \(x = - 2\)
\(y = {x^2} + 4x - 5\)
\(y = {( - 2)^2} + 4( - 2) - 5\)
\(= 4 - 8 - 5 = - 9\)
Therefore there is a minimum turning point at \(( - 2, - 9)\)
Cuts y-axis when \(x = 0\)
\(y = {x^2} + 4x - 5\)
\(y = {0^2} + 4(0) - 5 = - 5\)
Therefore cuts the \(y\)-axis at \((0, - 5)\)
Cuts \(x\)-axis when \(y = 0\)
\(y = {x^2} + 4x - 5\)
\({x^2} + 4x - 5 = 0\)
\((x + 5)(x - 1) = 0\)
\(x + 5 = 0\)
\(x = - 5\)
or
\(x - 1 = 0\)
\(x = 1\)
Therefore cuts the \(x\)-axis at \((-5, 0)\) and \((1, 0)\)
Question
Sketch the curve \(y = {x^3} + 3{x^2}\)
\(\frac{{dy}}{{dx}} = 3{x^2} + 6x\)
Stationary points occur when \(\frac{{dy}}{{dx}} = 0\).
\(3{x^2} + 6x = 0\)
\(3x(x + 2) = 0\)
\(3x = 0\)
\(x = 0\)
or
\(x + 2 = 0\)
\(x = -2\)
\(\frac{{dy}}{{dx}} = 3{( - 3)^2} + 6( - 3) = 27 - 18 = 9\) (positive)
\(\frac{{dy}}{{dx}} = 3{( - 2)^2} + 6( - 2) = 12 - 12 = 0\) (stationary)
\(\frac{{dy}}{{dx}} = 3{( - 1)^2} + 6( - 1) = 3 - 6 = - 3\) (negative)
\(\frac{{dy}}{{dx}} = 3{(0)^2} + 6(0) = 0 + 0 = 0\)(stationary)
\(\frac{{dy}}{{dx}} = 3{(1)^2} + 6(1) = 3 + 6 = 9\)(positive)
y-coordinates
When \(x = - 2\)
\(y = {x^3} + 3{x^2}\)
\(y = {( - 2)^3} + 3{( - 2)^2} = - 8 + 12 = 4\)
When \(x = 0\)
\(y = {x^3} + 3{x^2}\)
\(y = {0^3} + 3{(0)^2} = 0 + 0 = 0\)
Therefore there is a maximum turning point at \(( - 2, 4)\) and a minimum turning point at \((0,0)\)
Cuts y-axis when \(x = 0\)
\(y = {x^3} + 3{x^2}\)
\(y = {0^3} + 3{(0)^2} = 0\)
Therefore cuts the \(y\)-axis at \((0,0)\)
Cuts \(x\)-axis when \(y = 0\)
\(y = {x^3} + 3{x^2}\)
\({x^3} + 3{x^2} = 0\)
\({x^2}(x + 3) = 0\)
\({x^2} = 0\)
\(x = 0\)
or
\(x + 3 = 0\)
\(x = - 3\)
Therefore cuts the \(x\)-axis at \((0, 0)\) and \((- 3, 0)\)
