1) [tex]y=-0.5x-2[/tex]
The graph of the above equation is attached in the attachment below.
For all the values from negative infinity to -4, the function is positive.
X ∈ (-∞,-4]
2) [tex]y=-1.5x+3[/tex]
To find the x- intercept, plug y =0
[tex]0=-1.5x+3[/tex]
[tex]1.5x=3[/tex]
[tex]x=2[/tex]
x intercept = (2,0)
To find the y- intercept, plug x=0
[tex]y=-1.5(0)+3[/tex]
[tex]y=3[/tex]
Y- intercept = ( 0,3)
3) [tex]y=13-x[/tex]
Y- intercept = (0,13)
x- intercept = (13,0)
Area of triangle = [tex]\frac{bh}{2}[/tex]
Area of triangle = [tex]\frac{13\times13}{2}[/tex]
Area = [tex]\frac{169}{2}[/tex] sq unit.
4) The graphs of [tex]y=-2x+7[/tex] and [tex]y=0.5x-5.5[/tex] is attached in the attachment below.
The intersection point is ( 5,-3)
5) A = (-3,0)
B = (4,5)
C = (0,-4)
Slope of AB = [tex]\frac{5-0)}{4-(-3)} =\frac{5}{7}[/tex]
Slope intercept of line [tex]y=mx+b[/tex]
Where, m is the slope and b is the y- intercept.
Plugging point A and the slope in the y- intercept to find the value of b.
[tex]0=\frac{5(-3)}{7} +b[/tex]
[tex]b=\frac{15}{7}[/tex]
Equation of line AB: [tex]y=\frac{5x}{7} +\frac{15}{7}[/tex]
Slope of BC = [tex]\frac{-4-5}{0-4} =\frac{9}{4}[/tex]
Plug C = (0,-4)
[tex]-4=0+b[/tex]
b=-4
equation of line BC= [tex]y=\frac{9x}{4} -4[/tex]
Slope of line AC = [tex]\frac{-4-0}{0--3} =\frac{-4}{3}[/tex]
Equation of line AC = [tex]y=\frac{-4x}{3} -4[/tex]
Y intercept of AB = [tex](0,\frac{15}{7} )[/tex]
