Higher January 2020 Paper 1R Q18

EdexcelHigherCurrent spec7 marksDrawing & Using Graphs Linear Graphs & Gradients

18 The diagram shows the graph of \(\;y = \mathrm{f}(x)\;\) for \(\;-4 \leqslant x \leqslant 12\)

Graph of y = f(x) on a grid from x = −4 to 12 and y = −2 to 4, with a maximum at (0, 3), a minimum at (8, −1) and the point P on the curve at x = 2

The point \(P\) on the curve has \(x\) coordinate 2

(a) (i) Use the graph to find an estimate for the gradient of the curve at \(P\). (3)

(ii) Hence find an equation of the tangent to the curve at \(P\).
Give your answer in the form \(\;y = mx + c\) (2)

The equation \(\;\mathrm{f}(x) = k\;\) has exactly two different solutions for \(\;-4 \leqslant x \leqslant 12\)

(b) Use the graph to find the two possible values of \(k\). (2)

Mark scheme (a)(i)

Scheme	Marks
	B1
	M1
Working required Answer: −0.6	A1
	(3)

Notes

B1: tangent drawn at \(P\) (\(x = 2\))

M1: (dep on B1) for a method to find gradient e.g. \(\dfrac{\text{difference in } y\text{-values}}{\text{difference in } x\text{-values}}\)

A1: (dep on B1) accept answers in range −0.4 to −0.7 and from correct figures for their line

Mark scheme (a)(ii)

Scheme	Marks
e.g. \(y = -0.6x + c\) or \(y = mx + 3.6\) or \(2.4 = -0.6 \times 2 + c\)	M1
\(y = -0.6x + 3.6\)	A1
	(2)

Notes

M1: for start of method to find the tangent equation e.g. \(y = mx + c\) where \(m\) is their gradient or \(y = mx + c\) where \(c\) is the \(y\)-intercept for their tangent or for substituting a point from the curve e.g. (2, 2.4) into \(y = mx + c\) where \(m\) is their gradient

A1: ft their gradient from (i) and intercept of their tangent, so long as intercept / value of \(c\) is \(\gt 3\)

Scheme	Marks
3	B1
−1	B1
	(2)
	(7 marks)