Question

please help find the formulas for these!! ASAP!!!

Answer 1

Answer:

$a)\ f(x)=\left\{\begin{array}{cccc}-x+2&for&x\in(0,\ 2]&/0<x\leq2/\\\\-\dfrac{1}{3}x+\dfrac{5}{3}&for&x\in(2,\ 5]&/2<x\leq5/\end{array}\right$

$b)\ f(x)=\left\{\begin{array}{cccc}-3x-3&for&x\in(-1,\ 0]&/-1<x\leq0/\\\\-2x+3&for&x\in(0,\ 2]&/0<x\leq2/\end{array}\right$

Step-by-step explanation:

These are graphs of piecewise functions.

Each part is a graph of the linear function.

The slope-intercept form of an equation of a line:

$y=mx+b$

m - slope

b - y-intercept → (0, b)

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

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a)

Part 1:

Points (0, 2) → b = 2, and (2, 0).

Calculate the slope:

$m=\dfrac{0-2}{2-0}=\dfrac{-2}{2}=-1$

Put the value of m and b to the equation of a line:

$(1)\ y=-1x+2=-x+2$

Part 2:

Points (2, 1) and (5, 0).

Calculate the slope:

$m=\dfrac{0-1}{5-2}=\dfrac{-1}{3}=-\dfrac{1}{3}$

Put the value of the slope and the coordinates of the point (5, 0) to the equation of a line:

$0=-\dfrac{1}{3}(5)+b$

$0=-\dfrac{5}{3}+b$         add 5/3 to both sides

$\dfrac{5}{3}=b$

Therefore we have:

$y=-\dfrac{1}{3}x+\dfrac{5}{3}$

The domain of the function is (0, 2] and (2, 5].

Formula of the piecewise function:

$f(x)=\left\{\begin{array}{cccc}-x+2&for&x\in(0,\ 2]&/0<x\leq2/\\\\-\dfrac{1}{3}x+\dfrac{5}{3}&for&x\in(2,\ 5]&/2<x\leq5/\end{array}\right$

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b)

We solve the same as in a) in Part 1

Part 1:

(-1, 0), (0, -3) → b = -3

$m=\dfrac{-3-0}{0-(-1)}=\dfrac{-3}{1}=-3$

$y=-3x-3$

Part 2:

(0, 3) → b = 3, (2, -1)

$m=\dfrac{-1-3}{2-0}=\dfrac{-4}{2}=-2$

$y=-2x+3$

Domain: (-1, 0] and (0, 2].

$f(x)=\left\{\begin{array}{cccc}-3x-3&for&x\in(-1,\ 0]&/-1<x\leq0/\\\\-2x+3&for&x\in(0,\ 2]&/0<x\leq2/\end{array}\right$