Exercises: Linear functions, zeros, axis intercepts

$P(-25|30)$; $Q(55|-30)$

Determine the slope $m$ of the general line equation using the difference quotient .

$m=\frac{30-(-30)}{-25-55}=\frac{60}{-80}=-\frac{3}{4}$

Insert $m$ and the coordinates of a point, e.g. $P(-25|30)$, into the general line equation.

$30=\frac{-3}{4}(-25)+t$

Simplify: $\frac{-3}{4}(-25)=\frac{-3(-25)}{4}=\frac{3\cdot 25}{4}=\frac{75}{4}$

$\Rightarrow y=-\frac{3}{4}x+\frac{45}{4}$

Intersection with the $x$-axis

At the intersection with the $x$-axis the $y$-value is 0.

$0=\frac{-3}{4}x+\frac{45}{4}$

Solve for $x$. To do this, we use $|\cdot \frac{-4}{3}$

Note that both summands must be multiplied.

$0=\frac{(-3)(-4)}{4\cdot 3}x+\frac{45(-4)}{4\cdot 3}$

$\frac{(-3)(-4)}{4\cdot 3}=\frac{3\cdot 4}{12}=\frac{12}{12}=1$

$0=x+\frac{-180}{12}$

$0=x-15$

Add 15

$x=15$

$\Rightarrow$ The line intersects the $x$-axis at $S(15|0)$.