\(l{:}\,y = a x + b\) met \(a = \text{rc}_{l} = -5\)

Door \((0 , 4)\) dus \(b = 4 \text{,}\) en dus \(l{:}\,y = -5 x + 4\)

\(y = a x + b \text{.}\)

Door \((0 , -12) \text{,}\) dus \(b = -12 \text{.}\)

\(a = {\text{verticaal} \over \text{horizontaal}} = {-8 \over 10} = -\frac{4}{5} \text{.}\)

\(y = -\frac{4}{5} x - 12 \text{.}\)

De beginwaarde is \(b = 5 \text{.}\)

De verandering is \(a = 2 \text{.}\)

De gevraagde formule is dus \(K = 2 r + 5 \text{.}\)

\(l{:}\,y = a x + b\) met \(a = \text{rc}_{l} = \text{rc}_{m} = 3\)

Door \((0 , 5)\) dus \(b = 5 \text{,}\) en dus \(l{:}\,y = 3 x + 5\)

\(l{:}\,y = a x + b\) met \(a = \text{rc}_{l} = \text{rc}_{m} = -2\)

\(\begin{rcases}y = -2 x + b \\ \text{door } A (8 , 4)\end{rcases} \begin{matrix}-2 ⋅ 8 + b = 4 \\ -16 + b = 4 \\ b = 20\end{matrix}\)

Dus \(l{:}\,y = -2 x + 20\)

\(l{:}\,y = a x + b\) met \(a = \text{rc}_{l} = 3\)

\(\begin{rcases}y = 3 x + b \\ \text{door } A (7 , 9)\end{rcases} \begin{matrix}3 ⋅ 7 + b = 9 \\ 21 + b = 9 \\ b = -12\end{matrix}\)

Dus \(l{:}\,y = 3 x - 12\)

\(l{:}\,y = a x + b\) met \(a = {\Delta y \over \Delta x} = {-19 - -31 \over -2 - -4} = 6\)

\(\begin{rcases}y = 6 x + b \\ \text{door } A (-4 , -31)\end{rcases} \begin{matrix}6 ⋅ -4 + b = -31 \\ -24 + b = -31 \\ b = -7\end{matrix}\)

Dus \(l{:}\,y = 6 x - 7\)

\(y = a x + b\) met \(a = {\Delta y \over \Delta x} = {-27 - 15 \over 4 - -2} = -7\)

\(\begin{rcases}y = -7 x + b \\ \text{door } A (-2 , 15)\end{rcases} \begin{matrix}-7 ⋅ -2 + b = 15 \\ 14 + b = 15 \\ b = 1\end{matrix}\)

Dus \(y = -7 x + 1\)

\(l{:}\,y = a x + b\) met \(a = {\Delta y \over \Delta x} = {-8 - -8 \over 5 - -9} = {0 \over 14} = 0\)

\(\begin{rcases}y = b \\ \text{door } A (-9 , -8)\end{rcases} \begin{matrix}b = -8\end{matrix}\)

Dus \(l{:}\,y = -8\)

\(l{:}\,y = a x + b\) met \(a = {\Delta y \over \Delta x} = {-9 - -4 \over -8 - -8} = {-5 \over 0}\)

Delen door 0 is niet gedefinieerd, het is dus een verticale lijn.

Dus een verticale lijn met vergelijking \(l{:}\,x = -8\)

Door de oorsprong betekent dat \(b = 0 \text{,}\) dus \(l{:}\,y = a x\)

\(\begin{rcases}y = a x \\ \text{door } A (9 , 63)\end{rcases} \begin{matrix}a ⋅ 9 = 63 \\ a = 7\end{matrix}\)
Dus \(y = 7 x \text{.}\)

Evenredig betekent \(y = a x \text{.}\)

\(\begin{rcases}y = a x \\ \text{door } A (9 , 36)\end{rcases} \begin{matrix}a ⋅ 9 = 36 \\ a = 4\end{matrix}\)
Dus \(y = 4 x \text{.}\)

Rasterpunten \((5 , 10)\) en \((25 , 4)\) aflezen.

\(y = a x + b\) met \(a = {\Delta y \over \Delta x} = {4 - 10 \over 25 - 5} = -0{,}3\)

\(\begin{rcases}y = -0{,}3 x + b \\ \text{door } A (5 , 10)\end{rcases} \begin{matrix}-0{,}3 ⋅ 5 + b = 10 \\ -1{,}5 + b = 10 \\ b = 11{,}5\end{matrix}\)

Dus \(y = -0{,}3 x + 11{,}5\)

\(\begin{rcases}k \perp l \text{, dus } \text{rc}_{k} ⋅ \text{rc}_{l} = -1 \\ \text{rc}_{k} = 4\end{rcases} \text{rc}_{l} = -\frac{1}{4}\)

\(\begin{rcases}y = -\frac{1}{4} x + b \\ \text{door } A (-5 , 2)\end{rcases} \begin{matrix}2 = -\frac{1}{4} ⋅ -5 + b \\ 2 = 1\frac{1}{4} + b \\ b = \frac{3}{4}\end{matrix}\)
Dus \(l{:}\,y = -\frac{1}{4} x + \frac{3}{4} \text{.}\)

\({\Delta y \over \Delta x} = {19{,}59 - 18{,}48 \over 4 - 3} = 1{,}11\)

\({\Delta y \over \Delta x} = {24{,}03 - 19{,}59 \over 8 - 4} = 1{,}11\)
\({\Delta y \over \Delta x} = {30{,}69 - 24{,}03 \over 14 - 8} = 1{,}11\)

De gemiddelde verandering is steeds hetzelfde, dus de tabel hoort bij een lineair verband.

\(y = a x + b\) met \(a = 1{,}11\)

\(\begin{rcases}y = 1{,}11 x + b \\ x = 3 \text{ en } y = 18{,}48\end{rcases} \begin{matrix}1{,}11 ⋅ 3 + b = 18{,}48 \\ 3{,}33 + b = 18{,}48 \\ b = 15{,}15\end{matrix}\)

Dus \(y = 1{,}11 x + 15{,}15\)