Loav · VCE Mathematical Methods · Units 3 & 4
D03 · Differential calculus · Current curriculum
Connect a local slope to a line equation, and read a derivative graph as a record of how the original function changes.
1. A tangent needs a point and a gradient
For a differentiable function y = f(x), the tangent at x = a passes through (a, f(a)) and has gradient f′(a). The point-gradient form is therefore y − f(a) = f′(a)(x − a). Calculate the point from the original function and the gradient from the derivative. Confusing these two roles is one of the most common sources of tangent errors.
The tangent is a straight line that matches the curve’s value and first-order change at the contact point. It need not touch the curve only once, and it can cross the curve. For example, y = 0 is tangent to y = x³ at the origin even though the cubic crosses that line. The useful definition is local agreement in height and gradient, rather than an informal rule about touching without crossing.
A finite derivative gives a tangent of the form above. At a corner there is no single derivative, while a vertical tangent cannot be represented by a finite gradient. Check differentiability before substituting into the line formula. A displayed point must also lie in the stated function domain. For an endpoint of a contextual model, any tangent interpretation should be made from the side where that model is defined.
Worked example 1: a tangent to a cubic
Find the tangent to f(x) = x³ − 2x at x = 1. The contact point is (1, f(1)) = (1, −1). The derivative is f′(x) = 3x² − 2, so the gradient at that point is f′(1) = 1.
Substitute into point-gradient form: y + 1 = 1(x − 1), giving y = x − 2. To verify the line, substitute x = 1 and obtain y = −1, the curve’s value. Its gradient is also 1, the required derivative. These two checks catch many sign and intercept errors.
In fact f(x) − (x − 2) = x³ − 3x + 2 = (x − 1)²(x + 2). Thus the tangent meets the curve again at x = −2. A second intersection does not prevent tangency at x = 1; the tangent condition is about what happens locally at its contact point.
2. Normals and parallel tangents
A normal is the line through the contact point perpendicular to the tangent. If the tangent has finite non-zero gradient m, the normal has gradient −1/m because the product of perpendicular gradients is −1. In Worked example 1, the tangent gradient is 1, so the normal is y + 1 = −(x − 1), giving y = −x. It uses the same contact point as the tangent.
If the tangent is horizontal, its normal is vertical, with equation x = a. Do not divide by a zero tangent gradient. Conversely, a vertical tangent has a horizontal normal when that tangent is established geometrically. These special cases explain why a normal is best understood as a perpendicular line rather than just a reciprocal calculation. For a parallel tangent, the condition is instead equality of gradients.
A line y = mx + c has gradient m. To find points where a curve has a tangent parallel to that line, solve f′(a) = m. Then use each valid solution a to calculate f(a) and the associated tangent equation. The original line’s intercept is irrelevant to parallelism, although it matters if the problem asks for the tangent to pass through a specified point.
There can be zero, one or several qualifying contact points. If the curve is restricted to an interval, discard solutions outside that interval only after solving the derivative condition. Do not choose a single answer merely because the graph window shows one. A quadratic derivative can give two parallel tangents to a cubic, and a periodic derivative can give several within a bounded domain.
As a short illustration, for f(x) = x² − 4x + 1, a tangent parallel to a line of gradient 2 requires 2a − 4 = 2. Hence a = 3, f(3) = −2, and the tangent is y + 2 = 2(x − 3), or y = 2x − 8. The given comparison line and the tangent share a gradient, but generally have different intercepts.
3. A tangent through a point is not necessarily tangent there
When a tangent must pass through an external point, the point is usually not on the curve. Introduce a variable a for the unknown contact input. Write the tangent at a using f(a) and f′(a), then substitute the external point into that line. This produces an equation for a. Substituting the external point’s x-coordinate directly into the derivative would assume it is the contact point and generally solve the wrong problem.
Worked example 2: two tangents through one external point
Find the tangents to y = x² that pass through (0, −9). At an unknown contact input a, the point is (a, a²) and the gradient is 2a. The tangent is y − a² = 2a(x − a), or y = 2ax − a².
Because it passes through (0, −9), substitute that point: −9 = −a². Thus a² = 9 and a = ±3. For a = 3, the tangent is y = 6x − 9 and touches at (3, 9). For a = −3, it is y = −6x − 9 and touches at (−3, 9).
Both lines pass through the required external point. Both have gradients matching the derivative 2x at their own contact points. If the curve had instead been restricted to x ≥ 0, only the tangent at a = 3 would remain admissible. Always state all solutions consistent with the actual domain.
4. The derivative graph plots slopes, not heights
On the graph y = f′(x), the vertical coordinate at a is the tangent gradient of the original function at x = a. A value f′(a) = 4 means the original curve has gradient 4 there; it does not mean the original curve passes through (a, 4). The two graphs share their horizontal input, but the meanings of their vertical coordinates differ.
Where f′(x) > 0, the original function increases; where f′(x) < 0, it decreases. Where f′(x) = 0, the original function has a horizontal tangent if the derivative exists. A zero derivative can occur at a local maximum, a local minimum or a stationary point without a turn. To tell which, examine the derivative’s sign on the two sides rather than looking only at its value at the point.
A large positive derivative means steep upward change; a small positive derivative means gentler upward change. A derivative increasing from −5 to −1 still indicates that the original function is decreasing throughout, although its downward slopes are becoming less steep. “The derivative increases” and “the original function increases” are different statements. Read signs before reading the direction of the derivative graph.
5. Sketch a gradient graph systematically
Start with horizontal tangents in the original graph and mark corresponding zeros of the derivative. Then divide the x-axis into intervals where the original function rises or falls. Put the derivative above or below its axis accordingly. Compare steepness within each interval to decide roughly how the derivative values vary. Mark corners, jumps and other points where the derivative does not exist instead of drawing through them automatically.
For a straight segment with gradient m, the derivative is the horizontal line y = m on the open interval of that segment. For a constant segment, it is y = 0. At a join between straight segments with different slopes, the derivative is not defined even when the original graph is continuous. Endpoint values and open circles should reflect the domain and the nature of the join.
Worked example 3: connect a cubic to its gradient graph
Let f(x) = x³ − 3x. Its derivative is f′(x) = 3x² − 3 = 3(x − 1)(x + 1). This upward-opening parabola crosses the derivative’s x-axis at x = −1 and x = 1 and has value −3 at x = 0.
For x < −1 the derivative is positive, so f increases. Between −1 and 1 it is negative, so f decreases. For x > 1 it is positive again, so f increases. The original graph has a local maximum at (−1, 2) and a local minimum at (1, −2). These heights come from f, not from the derivative’s zero values.
At x = 0, the original curve passes through the origin with gradient −3. The derivative graph passes through (0, −3). Keeping those two points distinct is the central reading skill. A vertical translation f(x) + 7 has exactly the same derivative graph because adding a constant changes heights without changing slopes.
6. What a derivative graph does and does not determine
A derivative graph can determine intervals of increase and decrease, horizontal-tangent inputs and the gradients of tangents. On its own it does not determine the original function’s vertical position. The functions x², x² + 5 and x² − 8 all have derivative 2x. To reconstruct a particular member of such a family, extra information such as one function value is required.
When a question provides an approximate graph rather than an exact formula, do not invent exact coordinates that the scale does not support. Give qualitative intervals or suitable approximations, and distinguish clear structural information from estimated values. A neat sketch should show labelled axes, important input values, signs and any exclusions. It should not suggest precision beyond the supplied evidence.
7. Technology and final checks
In technology-active work, graph f and f′ together only if you label them unambiguously. Different vertical scales or an unsuitable viewing window can conceal derivative zeros and make a gentle slope look horizontal. Use algebra to confirm exact roots when possible. If a tangent is generated by software, check that its contact input is the requested one and that its equation satisfies both the point and gradient conditions.
For a final tangent answer, provide an equation rather than only a gradient. For a gradient-graph answer, explain which features establish its sign and zeros. For a parallel-tangent answer, verify every candidate against the domain. These short checks are more useful than a long calculation whose final line leaves the original geometric question unanswered.
Practice: seven original questions
- Find the tangent and normal to y = √x at x = 4.
- Find the tangent to y = ln x at x = e, leaving an exact answer.
- Find the tangent to y = x² + x that is parallel to y = 5x + 7.
- For f(x) = (x − 1)²(x + 2), find the zeros of f′ and the intervals where f is increasing.
- A derivative is f′(x) = (x − 2)(x + 1). Determine the increasing and decreasing intervals. Can you determine f(0) from this information alone?
- Find the tangent to y = eˣ at x = 0 and check that the contact point lies on it.
- A student says y = 0 cannot be tangent to y = x³ at the origin because the curve crosses the line. Evaluate the claim using a derivative.
Worked answers and reasoning
- The contact point is (4, 2). Since f′(x) = 1/(2√x), the gradient is 1/4. Hence y − 2 = (x − 4)/4, so y = x/4 + 1. The input 4 lies within the differentiable domain x > 0, and substituting it into the line returns 2. The normal gradient is −4, so y − 2 = −4(x − 4), or y = −4x + 18.
- The point is (e, 1), because ln e = 1. The gradient is 1/e. Thus y − 1 = (x − e)/e, giving y = x/e. This exact line contains (e, 1); replacing e prematurely by a rounded decimal would make the answer less precise.
- Parallelism requires derivative 2a + 1 = 5, so a = 2. The curve value is 2² + 2 = 6. Therefore y − 6 = 5(x − 2), or y = 5x − 4. Its gradient matches the comparison line, while its intercept need not match.
- Expand f(x) = x³ − 3x + 2, so f′(x) = 3x² − 3. Its zeros are x = −1 and 1. The derivative is positive for x < −1 and x > 1, so f is increasing on (−∞, −1) and (1, ∞). The derivative is negative between them.
- The factors give signs positive, negative, positive across the intervals cut by −1 and 2. Hence f increases on (−∞, −1) and (2, ∞), and decreases on (−1, 2). The value f(0) is undetermined because adding any constant to f preserves its derivative.
- The point is (0, 1) and the derivative eˣ has value 1 at zero. Thus y − 1 = x, or y = x + 1. Substituting x = 0 gives y = 1, confirming the contact point. The slope of the line also equals the curve’s derivative there.
- The claim is false. For f(x) = x³, f(0) = 0 and f′(0) = 3(0)² = 0. The tangent equation is y − 0 = 0(x − 0), namely y = 0. Crossing does not invalidate local agreement in value and gradient.
Curriculum and further study
Aligned to the current VCAA VCE Mathematics Study Design, Mathematical Methods Units 3 and 4, Calculus area of study and relevant outcome knowledge and skills. The VCAA Mathematical Methods FAQ, updated February 2026, clarifies inflection points and strict increasing or decreasing behaviour. Curriculum checked 15 October 2026. Unit 3 is the suggested teaching sequence; the official design specifies combined Units 3 and 4 coverage.
Further study: Evans, Lipson and Greenwood, Cambridge Senior Mathematics VCE: Mathematical Methods Units 3 & 4, second edition (2023), sections 9D and 10A. Independent Loav teaching material, not a VCAA or Cambridge publication. Explanations, worked examples, practice questions, answers and diagrams are original.
