Moderate -0.8 This is a straightforward application of differentiation to find a stationary point on a quadratic curve. The method is completely standard (differentiate, set to zero, verify minimum with second derivative), requires only basic calculus techniques, and the algebraic parameter 'a' doesn't add significant complexity since it remains constant throughout. Easier than average for AS-level.
Find the coordinates, in terms of \(a\), of the minimum point on the curve \(y = x^2 - 5x + a\), where \(a\) is a constant.
Fully justify your answer.
[3 marks]
Question 4:
4 | Selects an appropriate method –
either differentiates, at least as far
as:
dy
=2x....
dx
or commences completion of the
2
5
square:x− +...
2 | AO1.1a | M1 | 2
5 25
y = x− − +a
2 4
y minimised when squared bracket is 0
5 25
, a−
2 4
ALT
dy
=2x−5
dx
so 2x – 5 = 0 for minimum
5
x =
2
2
5 5 25
y = −5 +a=a−
2 2 4
Fully differentiates and sets
derivative equal to zero
or fully completes square
Allow one error | AO1.1a | M1
Obtains both coordinates | AO1.1b | A1
Total | 3
Q | Marking Instructions | AO | Marks | Typical Solution
Find the coordinates, in terms of $a$, of the minimum point on the curve $y = x^2 - 5x + a$, where $a$ is a constant.
Fully justify your answer.
[3 marks]
\hfill \mbox{\textit{AQA AS Paper 2 Q4 [3]}}