20N.1.SL.TZ0.S_5

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1 – (discriminant)

correct expression for $g$      (A1)

eg    $-\left(-x^2+4x+5\right)+k , x^2-4x-5+k=0$

evidence of discriminant      (M1)

eg    $b^2-4ac, \text{Δ}$

correct substitution into discriminant of $g$      (A1)

eg    ${\left(-4\right)}^2-4\left(1\right)\left(-5+k\right) , 16-4\left(k-5\right)$

recognizing discriminant is negative      (M1)

eg    $\text{Δ}<0 , {\left(-4\right)}^2-4\left(1\right)\left(-5+k\right)<0 , 16<4\left(k-5\right) , 16-4\left(-1\right)\left(5\right)<0$

correct working (must be correct inequality)      (A1)

eg    $-4k<-36 , k-5>4 , 16+20-4k<0$

$k>9$        A1 N3

 

METHOD 2 – (transformation of vertex of $\mathit{f}$)

valid approach for finding $f\left(x\right)$ vertex      (M1)

eg    $-\frac{b}{2a}=2 , f\text{'}\left(x\right)=0$

correct vertex of $f\left(x\right)$      (A1)

eg    $\left(2, 9\right)$

correct vertex of $-f\left(x\right)$      (A1)

eg    $\left(2, -9\right)$

correct vertex of $g\left(x\right)$      (A1)

eg    $\left(\begin{array}{c}2\\ -9\end{array}\right)+\left(\begin{array}{c}0\\ k\end{array}\right) , \left(2, -9+k\right)$

recognizing when vertex is above $x$-axis      (M1)

eg    $-9+k>0$, sketch

$k>9$        A1 N3

 

METHOD 3 – (transformation of $\mathit{f}$)

recognizing vertical reflection of $f\left(x\right)$      (M1)

eg    $-f\left(x\right) , x^2-4x-5$ , sketch

correct expression for $g\left(x\right)$      (A1)

eg    $x^2-4x-5+k$

valid approach for finding vertex of $g\left(x\right)$      (M1)

eg    $-\frac{b}{2a}=2 , g\text{'}\left(x\right)=0$

correct $y$ coordinate of vertex of $g\left(x\right)$      (A1)

eg    $y=-9+k , \left(2, -9+k\right)$

recognizing when vertex is above $x$-axis      (M1)

eg    $-9+k>0$ , sketch

$k>9$        A1 N3

 

[6 marks]

[N/A]