Further exercises: Linear functions

Draw the function $y=-3xy=-3x$

Mirror the line along the $xx$-axis.

Read the new equation of the function.

It is: $y=3xy=3x$

Move the straight line up by 2 units.

The $yy$-axis intercept has changed.

Read off the new function equation.

Answer: The straight line mirrored on the $xx$-axis and shifted upwards by $22$ units has the equation $y=3x+2y=3x+2$

Solution for b)

When mirroring on the $xx$-axis, the function term is multiplied by $(-1)(-1)$.

 $y=(-1)\cdot (-3x)=3xy=(-1)\backslash cdot(-3x)=3x$

Thus the equation of the mirrored straight line is $y=3x\text{}y=3x$

Move this straight line up by 3.

This corresponds to the $yy$-axis intercept $t\Rightarrow t\; \backslash Rightarrow$ $t=2t=2$

Answer: The new line equation is:

$y=3x+2y=3x+2$

As you can see from the picture, you have to move the line $hh$ in the $yy$-direction so that $ff$ and $gg$ then have the same zero.

$ff$ has the same slope as $hh$, i.e. $m=-1m=\; -1$.

Determine the zero of $gg$

Determine the line equation of $ff$

Insert the zero point $(-4\mathrm{\mid }0)(-4\; |\; 0)$ and the slope of $ff$ into the general line equation.

So the line equation for $ff$ is:

$f(x)=-x-4f(x)=-x-4$

For example, set $t=\{0;1;2;3;4\}t=\backslash left\backslash \{0;1;2;3;4\backslash right\backslash \}$ and calculate the corresponding $yy$-values. Other numbers for $tt$ are also admitted.

So we have the point $A(0\mathrm{\mid }3)A(0\backslash ;|\backslash ;3)$.

So we have the point $B(1\mathrm{\mid }3.5)B(1\backslash ;|\backslash ;3.5)$.

So we have the point $C(2\mathrm{\mid }4)C(2\backslash ;|\backslash ;4)$.

So we have the point $D(3\mathrm{\mid }4.5)D(3\backslash ;|\backslash ;4.5)$.

So we have the point $E(4\mathrm{\mid }5)E(4\backslash ;|\backslash ;5)$.

Draw the points A, B, C, D and E in a coordinate system.

$\Rightarrow \backslash ;\backslash ;\backslash Rightarrow\backslash ;\backslash ;$All points lie on a common line. This line has the equation $y=\frac{t}{2}+3y=\backslash frac\{t\}\{2\}+3$

Calculate when the $xx$ and $yy$ coordinates have the same value.

Set $yy$ equal to $tt$ for this.

$t=yt=y$

Replace $yy$ by the term ${\displaystyle \frac{t}{2}}+3\backslash dfrac\; t2+3$.

$\Rightarrow \backslash ;\backslash ;\backslash Rightarrow\backslash ;\backslash ;$The $xx$ - and $yy$ - coordinates have the same value when $t=6t=\; 6$. The corresponding point is $P\left(6\mathrm{\mid }6\right)P\backslash left(6\backslash ;|\backslash ;6\backslash right)$.

Plug $P_{1}P\_1$ and $P_{2}P\_2$ into the general line equation.

$y=m\cdot x+ty=m\backslash cdot\; x+t$

1) $k=m\cdot \sqrt{k}+tk=m\backslash cdot\backslash sqrt\{k\}+t$

2) $1=m\cdot 1+t1=m\backslash cdot1+t$

$1)-2):1)-2):$

Use the addition method.

$\Rightarrow \backslash Rightarrow$ The slope really is $a_1=\sqrt{k}+1a\_1=\backslash sqrt\{k\}+1$ and the line intersects the $yy$-axis at $P_y(0\mathrm{\mid }-\sqrt{k})\mathrm{.}P\_y(0\backslash ,|-\backslash sqrt\; k).$

Consider the set of points $P(\frac{k}{2}\cdot \sqrt{2}\mathrm{\mid }k)P(\backslash frac\{k\}\{2\}\backslash cdot\backslash sqrt\{2\}|k)$ for different $kk$.

To find the general equation of a straight line, we need two points on the straight line.

Therefore, it is best to choose here e.g. the two points $F(\frac{k}{2}\cdot \sqrt{2}\mathrm{\mid }k)F(\backslash frac\{k\}\{2\}\backslash cdot\backslash sqrt\{2\}|k)\backslash$and $G(\frac{k+1}{2}\cdot \sqrt{2}\mathrm{\mid }k+1)G(\backslash frac\{k+1\}\{2\}\backslash cdot\backslash sqrt\{2\}|k+1)$.

y$=m\cdot x+t=m\backslash cdot\; x+t$

First determine the slope $mm$.

${\displaystyle m=\frac{k+1-k}{\frac{k+1}{2}\cdot \sqrt{2}-\frac{k}{2}\cdot \sqrt{2}}=\frac{1}{\frac{\sqrt{2}}{2}}=\frac{2}{\sqrt{2}}}\backslash displaystyle\; m=\backslash frac\{k+1-k\}\{\backslash frac\; \{k+1\}\; \{2\}\; \backslash cdot\; \backslash sqrt\{2\}-\backslash frac\; k\; 2\; \backslash cdot\; \backslash sqrt\{2\}\}=\backslash frac\{1\}\{\backslash frac\{\backslash sqrt\{2\}\}\{2\}\}=\backslash frac\{2\}\{\backslash sqrt\{2\}\}$

Now you can insert $mm$ and one of the points, e.g. $PP$, into the general line equation.

$k=\underset{=k}{\underbrace{\frac{k}{2}\cdot \sqrt{2}\cdot \frac{2}{\sqrt{2}}}}+tk=\backslash underbrace\{\backslash frac\; k\; 2\; \backslash cdot\; \backslash sqrt\{2\}\; \backslash cdot\; \backslash frac\{2\}\{\backslash sqrt\{2\}\}\}\_\{=k\}+t$

Solve for $tt$.

$\Rightarrow \backslash Rightarrow$ t=0

Wrtite down the equation of the straight line.

$y=\frac{2}{\sqrt{2}}\cdot xy=\backslash frac\{2\}\{\backslash sqrt\{2\}\}\backslash cdot\; x$

Reading off intersection points