Part 1) we have
[tex]y=\frac{1}{2}x-4[/tex]
Statements
case A) The point [tex](-6,7)[/tex] is on the graph of the equation
The statement is false
we know that
If the point is on the graph of the equation
then
the point must be satisfy the equation
Verify
Substitute the value of x and y in the equation
[tex]7=\frac{1}{2}(-6)-4[/tex]
[tex]7=-7[/tex] ------> is not true
therefore
the point is not on the graph
case B) The point [tex](-4,-6)[/tex] is on the graph of the equation
The statement is true
we know that
If the point is on the graph of the equation
then
the point must be satisfy the equation
Verify
Substitute the value of x and y in the equation
[tex]-6=\frac{1}{2}(-4)-4[/tex]
[tex]-6=-6[/tex] ------> is true
therefore
the point is on the graph
case C) The graph of the equation is a single point representing one solution to the equation
The statement is false
Because, the graph of the equation is the set of all points that are solutions to the equation
case D) The graph of the equation is the set of all points that are solutions to the equation
The statement is true
The graph of the equation is the set of all points that satisfy the equation
Part 2) Graph the equation
[tex]4x-6y=48[/tex]
Find the intercepts of the equation
we know that
The x-intercept is the value of x when the value of y is equal to zero
The y-intercept is the value of y when the value of x is equal to zero
For [tex]x=0[/tex]
[tex]4*0-6y=48[/tex]
[tex]y=-8[/tex] -------> y-intercept [tex](0,-8)[/tex]
For [tex]y=0[/tex]
[tex]4x-6*0=48[/tex]
[tex]x=12[/tex] -------> x-intercept [tex](12,0)[/tex]
Plot the x-intercept and the y-intercept to graph the line
see the attached figure N [tex]1[/tex]
The answer Part 2) in the attached figure N [tex]1[/tex]
Part 3) Graph the function
[tex]f(x)=(2/3)x+1[/tex]
Find the intercepts of the equation
we know that
The x-intercept is the value of x when the value of y is equal to zero
The y-intercept is the value of y when the value of x is equal to zero
For [tex]x=0[/tex]
[tex]f(0)=(2/3)*0+1[/tex]
[tex]f(0)=1[/tex] -------> y-intercept [tex](0,1)[/tex]
For [tex]f(x)=0[/tex]
[tex]0=(2/3)x+1[/tex]
[tex]x=-1.5[/tex] -------> x-intercept [tex](-1.5,0)[/tex]
Plot the x-intercept and the y-intercept to graph the line
see the attached figure N [tex]2[/tex]
The answer Part 3) in the attached figure N [tex]2[/tex]
Part 4) Which answer is an equation in point-slope form for the given point and slope?
point: [tex](-3,5)[/tex] ; slope: [tex]4[/tex]
we know that
the equation of the line in point-slope form is equal to
[tex]y-y1=m(x-x1)[/tex]
Substitute the values
[tex]y-5=4(x+3)[/tex]
therefore
The answer part 4) is the option C
[tex]y-5=4(x+3)[/tex]
Part 5) Which ordered pairs are on the line [tex]4x-3y=8[/tex] ?
we know that
If a point is on the line
then
the point must be satisfy the equation of the line
we're going to verify every case
case A) [tex](3/2,-2/3)[/tex]
Substitute the value of x and y in the equation of the line
[tex]4*(3/2)-3*(-2/3)=8[/tex]
[tex]6+2=8[/tex]
[tex]8=8[/tex] -------> is true
therefore
the point [tex](3/2,-2/3)[/tex] is on the line
case B) [tex](-1,4)[/tex]
Substitute the value of x and y in the equation of the line
[tex]4*(-1)-3*(4)=8[/tex]
[tex]-4-12=8[/tex]
[tex]-16=8[/tex] -------> is not true
therefore
the point [tex](-1,4)[/tex] is not on the line
case C) [tex](-4,-8)[/tex]
Substitute the value of x and y in the equation of the line
[tex]4*(-4)-3*(-8)=8[/tex]
[tex]-16+24=8[/tex]
[tex]8=8[/tex] -------> is true
therefore
the point [tex](-4,-8)[/tex] is on the line
case D) [tex](1,-4/3)[/tex]
Substitute the value of x and y in the equation of the line
[tex]4*(1)-3*(-4/3)=8[/tex]
[tex]4+4=8[/tex]
[tex]8=8[/tex] -------> is true
therefore
the point [tex](1,-4/3)[/tex] is on the line
therefore
the answer Part 5) is
[tex](3/2,-2/3)[/tex]
[tex](-4,-8)[/tex]
[tex](1,-4/3)[/tex]
