IB Maths AI SL 2 Functions

$\frac{9.45-8.73}{1958-1708}$

the (mean) yearly change in (mean annual) temperature

Note: Accept equivalent statements, e.g. "rate of change of temperature".

$\quad{ }^{\circ} \mathrm{C} /$ year OR degrees C per year A1

Note: Do not follow through from part (b)(i) into (b)(ii).

attempt to substitute point and gradient into appropriate formula $8.73=0.00288 \times 1708+c \Rightarrow c=3.81096 \ldots$
or $9.45=0.00288 \times 1958+c \Rightarrow c=3.81096$.
equation is y=0.00288 x+3.81

attempt to substitute 2000 into their part (c)
$0.00288 \times 2000+3.81096 \ldots$$=9.57\left({ }^{\circ} \mathrm{C}\right)(9.57096 \ldots)$

cannot (always reliably) make a prediction of x from a value of y, when using a y on x line / regression line is not x on y
extrapolation

$-2.6\left(\frac{-13}{5}\right)$

EITHER
$(-\infty,-3.67] \cup[-2.6, \infty)$
A1A1A1
OR
$y \leq-3.67$ or $y \geq-2.6$
A1A1A1

Note: Award A1 for $-3.67\left(-3.66666 \ldots,-\frac{11}{3}\right)$ seen
A1 for their $-2.6\left(-\frac{13}{5}\right)$ seen and both sets of inequalities/brackets correct;
A1 if completely correct, including union symbol or "or".
It is possible to award A0A1A0 for correctly presenting their -3.67 (and their -2.6 ).
The -3.67 maybe seen as part of a sketch.

Award A1A0A0 for $-3.67 \leq y \leq-2.6$ or [-3.67,-2.6].

setting f(x) equal to -2 f(x)=-2 OR $-2=\frac{4}{x}-3$ OR sketch of graph with line labelled -2

$\quad\left(b=\frac{360}{15}=\right) 24$

$\frac{\max +\min }{2}$

$h(t)=18.5 \sin (24 t)+26.5$

Note: Award A1 for 18.5 seen as parameter a, A1 for a completely correct equation (including LHS).

attempt to equate 33 with h(t) OR sketch graph of curve and line
$h(t)(=18.5 \sin (24 t)+26.5)=33$$t=0.857(0.857083 \ldots)$ and $t=6.64(6.64291 \ldots)$ (minutes)