Question

On a coordinate plane, a solid straight line has a positive slope and goes through (negative 4, 1) and (0, 3). Everything below and to the right of the line is shaded.
Which linear inequality is represented by the graph?

y ≤ 2x + 4
y ≤ 1/2x + 3
y ≥ 1/2x + 3
y ≥ 2x + 3

Answer 1

Answer:

y ≤ $\frac{1}{2} x+3$

Step-by-step explanation:

Since everything below and to the right of the line is shaded, we must have the following:

y ≤

because the symbol  ≤ implies that the area below the line is the shaded part

with this, options 3 and 4 are discarded

now, on the right side of the symbol we should have the expression for a line, which as a general form:

$mx+b$

where m is the slope and b is the y-intercept of the line

so until now the answer should be in the form

y ≤ $mx+b$

and we calculate the slope m with the two points we are given:

(-4, 1) and (0, 3)

where:

$x_{1}=-4\\y_{1}=1\\x_{2}=0\\y_{2}=3$

and we plug this values in the slope equation:

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\\\m=\frac{3-1}{0-(-4)}\\ \\m=\frac{2}{4}\\ \\m=\frac{1}{2}$

so now we know that our solution must have the form:

y ≤ $\frac{1}{2} x+b$

and b can be found since we know that the line passes through (0, 3), so when x=0 y=3, this means that the y-intercept b is 3:

y ≤ $\frac{1}{2} x+3$

which is the second option

Answer 2

Answer:

y ≥ 2x + 3

Step-by-step explanation: