Question

On a coordinate plane, a piecewise function has 2 lines. The first line has a closed circle at (0, 3) and then goes down through (negative 4, 1) with an arrow instead of an endpoint. The second line has an open circle at (0, 1) and then goes down through (4, negative 1) with an arrow instead of an endpoint.
Consider the function given by the graph. What are these values?

f(–2 ) =

f(0) =

f(4) =

Answer 1

Answer:

f(-2) = 2

f(0) = 3

f (4) = -1

Step-by-step explanation:

The equation of the straight line passing through the points (0,3) and (-4,1) is  

$\frac{y - 1}{1 - 3} = \frac{x - (- 4)}{-4 - 0}$

⇒ 2(y - 1) = x + 4

⇒ 2y - 2 = x + 4

⇒ 2y = x + 6

⇒ $y = \frac{1}{2} x + 3$

Again, the equation of the line passing through the points (0,1) and (4,-1) is

$\frac{y - (- 1)}{- 1 - 1} = \frac{x - 4}{4 - 0}$

⇒ - 2 ( y + 1) = x - 4

⇒ - 2y = x - 2

⇒ $y = 1 - \frac{1}{2}x$

Therefore, the function is defined as

$f(x) = \frac{1}{2} x + 3$ for x ≤ 0 ........... (1) and  

$f(x) = 1 - \frac{1}{2}x$ for x > 0 ............ (2)

Therefore, f(-2) = 2 from equation (1).

f(0) = 3 from equation (1)

And f (4) = -1 from equation (2). (Answer)

Answer 2

Answer:

f(-2) = 2

f(0) = 3

f(4) = -1

Step-by-step explanation:

The equation of the line passing through (0,3) and (-4,1) will be

y-y_{1} =  m\times (x-x_{1}  )

m = \frac{3-1}{0+4}

therefore

y - 3 = \frac{1}{2}\times x

therefore f(-2) = (\frac{1}{2}\times -2) + 3 = 2

since we have a closed circle at (0,3)

the function value at (0,3) will be 3

therefore f(0) = 3

The equatiom of the line passing through (0, 1) and (4,-1) will be

y-y_{1} =  m\times (x-x_{1}  )

m = \frac{1+1}{0-4}

therefore

y - 1 = \frac{-x}{2}

therefore f(4) = \frac{-4}{2} + 1 = -1 .