| Date | November 2007 | Marks available | 3 | Reference code | 07N.2.sl.TZ0.5 |
| Level | SL only | Paper | 2 | Time zone | TZ0 |
| Command term | Find | Question number | 5 | Adapted from | N/A |
Question
The diagram below shows the graph of a line \(L\) passing through (1, 1) and (2 , 3) and the graph \(P\) of the function \(f (x) = x^2 − 3x − 4\)
Find the gradient of the line L.
Differentiate \(f (x)\) .
Find the coordinates of the point where the tangent to P is parallel to the line L.
Find the coordinates of the point where the tangent to P is perpendicular to the line L.
Find
(i) the gradient of the tangent to P at the point with coordinates (2, − 6).
(ii) the equation of the tangent to P at this point.
State the equation of the axis of symmetry of P.
Find the coordinates of the vertex of P and state the gradient of the curve at this point.
Markscheme
for attempt at substituted \(\frac{{ydistance}}{{xdistance}}\) (M1)
gradient = 2 (A1)(G2)
[2 marks]
\(2x - 3\) (A1)(A1)
(A1) for \(2x\) , (A1) for \(-3\)
[2 marks]
for their \(2x - 3 =\) their gradient and attempt to solve (M1)
\(x = 2.5\) (A1)(ft)
\(y = -5.25\) ((ft) from their x value) (A1)(ft)(G2)
[3 marks]
for seeing \(\frac{{ - 1}}{{their(a)}}\) (M1)
solving \(2x – 3 = - \frac{1}{2}\) (or their value) (M1)
x = 1.25 (A1)(ft)(G1)
y = – 6.1875 (A1)(ft)(G1)
[4 marks]
(i) \(2 \times 2 - 3 = 1\) ((ft) from (b)) (A1)(ft)(G1)
(ii) \(y = mx + c\) or equivalent method to find \(c \Rightarrow -6 = 2 + c\) (M1)
\(y = x - 8\) (A1)(ft)(G2)
[3 marks]
\(x = 1.5\) (A1)
[1 mark]
for substituting their answer to part (f) into the equation of the parabola (1.5, −6.25) accept x = 1.5, y = −6.25 (M1)(A1)(ft)(G2)
gradient is zero (accept \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\)) (A1)
[3 marks]
Examiners report
Parts (a) and (b) were very well done. After that, only the stronger candidates were able to cope. The equation of the tangent at the point with coordinates (2, 6) was badly done but some candidates managed to find the equation of the tangent line from their GDC. The equation of the axis of symmetry was reasonably well done although many just wrote down 1.5 instead of x = 1.5.
Some forgot to write down that the gradient at the vertex was 0.
Parts (a) and (b) were very well done. After that, only the stronger candidates were able to cope. The equation of the tangent at the point with coordinates (2, 6) was badly done but some candidates managed to find the equation of the tangent line from their GDC. The equation of the axis of symmetry was reasonably well done although many just wrote down 1.5 instead of x = 1.5.
Some forgot to write down that the gradient at the vertex was 0.
Parts (a) and (b) were very well done. After that, only the stronger candidates were able to cope. The equation of the tangent at the point with coordinates (2, 6) was badly done but some candidates managed to find the equation of the tangent line from their GDC. The equation of the axis of symmetry was reasonably well done although many just wrote down 1.5 instead of x = 1.5.
Some forgot to write down that the gradient at the vertex was 0.
Parts (a) and (b) were very well done. After that, only the stronger candidates were able to cope. The equation of the tangent at the point with coordinates (2, 6) was badly done but some candidates managed to find the equation of the tangent line from their GDC. The equation of the axis of symmetry was reasonably well done although many just wrote down 1.5 instead of x = 1.5.
Some forgot to write down that the gradient at the vertex was 0.
Parts (a) and (b) were very well done. After that, only the stronger candidates were able to cope. The equation of the tangent at the point with coordinates (2, 6) was badly done but some candidates managed to find the equation of the tangent line from their GDC. The equation of the axis of symmetry was reasonably well done although many just wrote down 1.5 instead of x = 1.5.
Some forgot to write down that the gradient at the vertex was 0.
Parts (a) and (b) were very well done. After that, only the stronger candidates were able to cope. The equation of the tangent at the point with coordinates (2, 6) was badly done but some candidates managed to find the equation of the tangent line from their GDC. The equation of the axis of symmetry was reasonably well done although many just wrote down 1.5 instead of x = 1.5.
Some forgot to write down that the gradient at the vertex was 0.
Parts (a) and (b) were very well done. After that, only the stronger candidates were able to cope. The equation of the tangent at the point with coordinates (2, 6) was badly done but some candidates managed to find the equation of the tangent line from their GDC. The equation of the axis of symmetry was reasonably well done although many just wrote down 1.5 instead of x = 1.5.
Some forgot to write down that the gradient at the vertex was 0.
