Goal and prerequisites
Differentiate simple polynomials, build a tangent and distinguish a minimum from a maximum. Review coordinate geometry before starting.
Method and its conditions
For axⁿ use derivative anx^(n−1), with constant derivative zero. Differentiate term by term. Substitute an input for a gradient; solve dy/dx=0 for stationary inputs, then substitute into the original function for their heights.
Worked example: show the decisions
For y=x²−6 x+4, the derivative is 2 x−6. It vanishes at x=3; the original function gives y=−5. The derivative is negative before 3 and positive after 3, confirming a minimum. For y=2 x²+1 at x=2, slope is 8 and the point is(2,9), so the tangent is y=8 x−7.
Contrast: diagnose an incorrect approach
Setting the original function equal to zero finds roots, not stationary points. Differentiation gives a new function for gradient, not the original height. Keep these roles distinct when a question asks for a tangent intercept or a turning-point coordinate.
Transfer: try a changed situation
For y=−x²+8 x+3, find the stationary point and justify its type.
Transfer solution and interpretation
The derivative −2 x+8 vanishes at x=4; substitution gives y=19. The derivative changes from positive to negative, so (4,19) is a maximum. This stays within the 0580 requirement for simple polynomial derivatives.
y=3 x³+5. Find dy/dx at x=2.
- Differentiate the cubic term using the power rule; the constant differentiates to zero.
- Using the stated quantities: 3×3×2^2 = 36.
- Check the domain, units and whether the answer is an exact value, bound or estimate.
Try it, then explain it.
Complete a short set without looking at the example. After each answer, explain which operation preserved the relationship and why the result makes sense.
Sources & authorship
Original Slope explanations and examples. References support the concepts; they are not the source of copied questions. Independent educator review is pending.
0580 syllabus for exams in 2025–2027