Standard +0.3 This is a straightforward application of the quotient rule followed by solving a quadratic equation. The question requires finding dy/dx using the quotient rule, setting it equal to -4, and solving for x-coordinates, then finding corresponding y-values. While it involves multiple steps, each step is routine and the problem type (finding points with given gradient) is a standard textbook exercise with no novel insight required.
Use quotient rule or, after adjustment, product rule. Obtain \(\frac{3x - 15 - 3x - 1}{(x - 5)^2}\) or equivalent
M1* A1
Equate first derivative to \(-4\) and solve for \(x\). Obtain x-coordinates 3 and 7 or one correct pair of coordinates. Obtain y-coordinates \(-5\) and \(11\) respectively or other correct pair of coordinates
M1 dep A1 A1
[5]
Use quotient rule or, after adjustment, product rule. Obtain $\frac{3x - 15 - 3x - 1}{(x - 5)^2}$ or equivalent | M1* A1 |
Equate first derivative to $-4$ and solve for $x$. Obtain x-coordinates 3 and 7 or one correct pair of coordinates. Obtain y-coordinates $-5$ and $11$ respectively or other correct pair of coordinates | M1 dep A1 A1 | [5]
2 A curve has equation
$$y = \frac { 3 x + 1 } { x - 5 }$$
Find the coordinates of the points on the curve at which the gradient is - 4 .
\hfill \mbox{\textit{CAIE P2 2015 Q2 [5]}}