Linear relations > Linear functions

Exercise 1

a

Do it.

b

Cyclist 1: ${a}_{1}=20t$

Cyclist 2: ${a}_{2}=-25t+150$

c

$20t=-25t+150$ gives you $45t=150$ and therefore $t=3\frac{1}{3}$.

The cyclists pass each other after $3$ hours and $20$ minutes.

Exercise 2

a

$3x-5=0$ therefore $3x=5$ and thus $x=\frac{5}{3}$.The axis intersects are: $(\frac{5}{3},0)$ and $(0,-5)$.

b

$x-4=0$ therefore $x=4$. The axis intersects are: $(4,0)$ and $(0,-4)$.

c

$-0.5x+4=0$ therefore $0.5x=4$ and thus $x=8$. The axis intersects are: $(8,0)$ and $(0,4)$.

d

$-2(x+3)=0$ therefore $x+3=0$ and thus $x=-3$. The axis intersects are: $(-3,0)$ and $(0,-6)$.

Exercise 3

a

y-intersect = $(0,-4)$, and slope $5$.

b

${y}_{2}=6+5x$

c

${y}_{3}=-4-5x$

Exercise 4

a

Linear equation $y=ax+b$ through $(1,5)$ gives you $5=a+b$, therefore $b=5-a$. Substitution results in $y=ax+5-a$.

b

If the line goes through $A$, then $a=-4$ and if the line goes through $C$, then $a=-\frac{2}{3}$. Therefore the line will not intersect the square if $a<-4\vee a>-\frac{2}{3}$.

Exercise 5

a

$f$: $4-0.5x=0$ therefore $0.5x=4$ and thus $x=8$. The axis intersects are: $(8,0)$ and $(0,4)$.

$g$: $2x-1=0$ therefore $2x=1$ and thus $x=\frac{1}{2}$. The axis intersects are: $(\frac{1}{2},0)$ and $(0,-1)$.

b

$4-0.5x=2x-1$ gives you $-2.5x=-5$ and therefore $x=2$. The intersect is $(2,3)$.

Exercise 6

a

$26$ km in $45$ minutes gives you $34\frac{2}{3}$ km/h.

b

The participant starts the last part of the tour at $t=7$, with $a=174$, and has finished the tour at $t=7\frac{3}{4}$, with $a=200$.

Therefore $a\left(t\right)=34\frac{2}{3}t+b$. Using $t=7$ you get $174=242\frac{2}{3}+b$, and thus $b=-68\frac{2}{3}$. The function rule is $a\left(t\right)=34\frac{2}{3}t-68\frac{2}{3}$.

c

At $t=9$, $a=174$ and at $t=10$, $a=200$. Therefore $a\left(t\right)=26t+b$ and $200=260+b$, so $b=-60$. The function rule is $a\left(t\right)=26t-60$.