Unit 8
7 marks · EasyThis question is about using a graph to solve a quadratic equation and expressing a vector as a column vector.
The sketch below shows the graph of for .
Write down the values of where the graph crosses the -axis.
By drawing a suitable horizontal line on the graph, explain how you would use it to solve .
A triangle has vertices at and .
Write down the vector as a column vector.
The vector is given by .
What are the components of ?
The vector .
Write down the magnitude of .
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a(i)) x = 1 and x = 3
(a(ii)) Draw the line y = 1 and find x-coordinates where it meets the curve
(b)
(c)
(d) 5
Show mark scheme and worked solution
Marking points
B1 (a(i)) · 1 mark · States the correct x-values, x = 1 and x = 3; 1 for both or FT from graph. (follow-through)
M1 (a(ii)) · 1 mark · Communicates that the horizontal line y = 1 should be drawn/used.
A1 (a(ii)) · 1 mark · Describes that the x-coordinates where this line meets the curve are solutions to x^2 - 4x + 1 = 0. (follow-through)
B1 (b) · 1 mark · Writes \overrightarrow{AB} = (3, 4)^T.
B1 (c) · 1 mark · Finds correct vector .
M1 (d) · 1 mark · Uses correct method for magnitude (Pythagoras theorem).
A1 (d) · 1 mark · Gives magnitude as 5. (follow-through)
Worked solution
a(i) The graph crosses the x-axis when y = 0.
From the table of points or the graph, this happens at x = 1 and x = 3.
a(ii) To solve x^2 - 4x + 1 = 0 graphically, draw the horizontal line y = 1 on the graph. The solutions are the x-values where this line meets the curve y = x^2 - 4x + 3.
b The vector = (4 - 1, 6 - 2) = (3, 4). Column vector: .
c .
d The magnitude of is .