99 points for whoever can answer these! show work! 1) what is the slope of the line through the points (–3, 5) and (4, 5)? 2) write an equation in slope-intercept form that represents a line with slope -2 through the point (2, 3). 3) write an equation in standard form that represents a line with slope 5 through the point (0, 2). 4) answer the following questions for the equation: y = -3x +5+7 what direction does the graph open? state the vertex as an ordered pair.

1. The slope is the y-difference divided by the x-difference:
.. m = (5 -5)/(4 -(-3)) = 0
The slope of the line is 0.

2. In point-slope form, the equation is
.. y = -2(x -2) +3
Expanding to slope-intercept form, you have
.. y = -2x +7

3. You can start with point-slope form
.. y = 5(x -0) +2
.. 5x -y = -2 . . . . . standard form

4) The questions seem to relate to a parabola. The given equation is a linear equation. Perhaps you mean
.. y = -3x² +5x +7
a) the negative coefficient of x^2 tells you the graph opens downward.

b) The vertex x-coordinate is -5/(2*(-3)) = 5/6. The y-value of the vertex is
.. (-3*5/6 +5)*5/6 +7 = 5/2*5/6 +7 = 9 1/12
As an ordered pair, the vertex is (x, y) = (5/6, 9 1/12)

1) What is the slope of the line through the points (–3, 5) and (4, 5)?
P1=(-3,5)=(x1,y1)→x1=-3, y1=5
P2=(4,5)=(x2,y2)→x2=4, y2=5
Slope: m=(y2-y1) / (x2-x1)
m=(5-5) / (4-(-3))
m= (0) / (4+3)
m= (0) / (7)
m=0
The slope of the line through the points (-3,5) and (4,5) is 0.

2) Write an equation in slope-intercept form that represents a line with slope -2 through the point (2, 3).
Slope: m=-2
Point P1=(2,3)=(x1,y1)→x1=2, y1=3
y-y1=m(x-x1)
y-3=-2(x-2)
y-3=-2x+4
y-3+3=-2x+4+3
y=-2x+7
Equation in slope-intercept form that represents a line with slope -2 through the point (2,3) is y=-2x+7

3) Write an equation in standard form that represents a line with slope 5 through the point (0, 2).
Slope: m=5
Point P1=(0,2)=(x1,y1)→x1=0, y1=2
y-y1=m(x-x1)
y-2=5(x-0)
y-2=5x
y-2-y=5x-y
-2=5x-y
5x-y=-2
The equation in standard form that represents a line with slope 5 through the point (0, 2) is 5x-y=-2

4) Answer the following questions for the equation: y = -3x^2+5x+7 What direction does the graph open? State the vertex as an ordered pair.
y=ax^2+bx+c
a=-3, b=5, c=7
a=-3<0, then the graph opens downward
Vertex: V=(xv,yv)=(h,k)
xv=h=-b/(2a)
h=-5 / [2(-3)]
h= -5 / (-6)
h=5/6

yv=k=-3(5/6)^2+5(5/6)+7
k=-3 [ (5)^2 / (6)^2 ]+(5*5)/6+7
k=-3(25/36)+25/6+7
k=-25/12+25/6+7
k=(-25+2*25+12*7) / 12
k=(-25+50+84) / 12
k=109/12
Vertex: V=(5/6,109/12)