(a) Write \((2x - 5)(x + 4)\) in the form \(2(x + a)^2 - b\).
You must show your working. [5]
(b) Charlie, Dev and Eve all attempt to sketch the graph of \(y = (2x - 5)(x + 4)\).
Whose sketch is the most accurate? Write down the properties of the graph that you used in making your decision. [2]
Mark scheme (a)
Answer
Marks
Part marks and guidance
\(2\left(x + \frac{3}{4}\right)^2 - \frac{169}{8}\) as final answer with correct working
5
Method 1: B3 for \(2\left(x + \frac{3}{4}\right)^2\) in final answer with correct working or M1 for \(2x^2 - 5x + 8x - 20\) oe M1 for \(2\left(x^2 + \frac{3}{2}x\right)\) [– 20] oe
AND
M1 for [\(-b\) =] \(-2\left(their\ \frac{3}{4}\right)^2 - 20\) soi by \(-\frac{169}{8}\)
Method 2: B3 for \(2\left(x + \frac{3}{4}\right)^2\) in final answer with correct working or M1 for \(2x^2 - 5x + 8x - 20\) oe or for \(2(x^2 + ax + ax + a^2) - b\) oe M1 for \(4ax = 3x\) soi by [\(a\) =] \(\frac{3}{4}\)
AND
M1 for [\(-b\) =] \(-2\left(their\ \frac{3}{4}\right)^2 - 20\) soi by \(-\frac{169}{8}\)
Method 3: B3 for \(2\left(x + \frac{3}{4}\right)^2\) in final answer with correct working or M1 for roots \(-4\) and 2.5 M1 for turning point at [\(x\) =] \(\frac{-4 + 2.5}{2}\) soi by \(-\frac{3}{4}\)
AND
M1 for [\(-b\) =] \(\left(2\left(their\ -\frac{3}{4}\right) - 5\right)\left(\left(their\ -\frac{3}{4}\right) + 4\right)\) soi by \(-\frac{169}{8}\)
If no or insufficient working SC2 for \(2\left(x + \frac{3}{4}\right)^2 - \frac{169}{8}\) or SC1 for \(2\left(x + \frac{3}{4}\right)^2\) [\(+k\)]
‘Correct working’ requires evidence of at least M1 Accept decimal and mixed number equivalents throughout eg. \(2(x + 0.75)^2 - 21.125\) \(2\left(x + \frac{3}{4}\right)^2 - 21\frac{1}{8}\)
May be in a grid (Methods 1 and 2)
Mark scheme (b)
Answer
Marks
Part marks and guidance
Charlie with at least one bullet point and no incorrect statements:
The roots are \(-4\) and 2.5
The turning point is at [\(x\) =] \(-\frac{3}{4}\) oe and only one root is positive/negative
The turning point is at [\(x\) =] \(-\frac{3}{4}\) oe and \(y\)-intercept is \(-20\) or negative
2
B1 for Charlie with one bullet point and no incorrect statements:
Turning point is at [\(x\) =] \(-\frac{3}{4}\) oe
\(y\)-intercept is \(-20\) or negative
or SC1 for any of the following with no incorrect statements: Dev and \(y\)-intercept is \(-20\) or negative or Eve and turning point is at [\(x\) =] \(-\frac{3}{4}\) oe or A person linked correctly to a FT turning point from (a)
