Question

1. a linear function (f(x)) contains the points (( - 3,6)) and ((6,0)). (a) determine an equation for (f(x)) in slope - intercept form. (b) does the graph of the function pass through the point ((30,-12))? justify your response. 8. a line has a (y) - intercept of 16 and an (x) - intercept of 4. another line is perpendicular to this line and has the same (x) - intercept of 4. what is the (y) - intercept of the second line? show how you found your answer.

Explanation:

(a)

Step1: Calculate the slope

The slope $m$ of a line passing through two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by $m=\frac{y_2 - y_1}{x_2 - x_1}$. Here, $(x_1,y_1)=(-3,6)$ and $(x_2,y_2)=(6,0)$. So $m=\frac{0 - 6}{6-(-3)}=\frac{-6}{9}=-\frac{2}{3}$.

Step2: Find the y - intercept

The slope - intercept form of a line is $y = mx + b$. We can use one of the points, say $(6,0)$, and the slope $m =-\frac{2}{3}$ to find $b$. Substitute $x = 6$, $y = 0$ and $m=-\frac{2}{3}$ into $y=mx + b$: $0=-\frac{2}{3}\times6 + b$. Simplify the right - hand side: $0=-4 + b$, so $b = 4$.
The equation of the function $f(x)$ in slope - intercept form is $y=-\frac{2}{3}x + 4$.

Step1: Substitute the point into the equation

We have the equation $y=-\frac{2}{3}x + 4$. Substitute $x = 30$ and $y=-12$ into it. The left - hand side is $y=-12$, and the right - hand side is $-\frac{2}{3}\times30+4=-20 + 4=-16$.
Since $-12
eq-16$, the point $(30,-12)$ does not lie on the graph of the function.

Answer:

$y=-\frac{2}{3}x + 4$

(b)