help please

josh123

Junior Member
Joined
Oct 5, 2005
Messages
52
Hey guys I need some help. Can someone show me how to do these problems?


Line thru 2 pointsAn equation for a line passing thru (1, 2) and (4, 6) is

a. 4x - 3y + 2 = 0
b. 4x + 3y + 2 = 0
c. 4x - 3y - 2 = 0
d. 4x + 3y - 2 = 0
e. none of these


Cartesian plane and graphing 3
The only quadrant which contains no points of the graph of y = -x2 - 8x - 14 is quadrant

a. I
b. II
c. III
d. IV
e. There are points on this graph in all four quadrants.


Thanks,
Josh
 
josh123 said:


Line thru 2 pointsAn equation for a line passing thru (1, 2) and (4, 6) is

a. 4x - 3y + 2 = 0
b. 4x + 3y + 2 = 0
c. 4x - 3y - 2 = 0
d. 4x + 3y - 2 = 0
e. none of these
Ignoring multi-choice and applying some learning instead:
- find the slope of the line, given that you know two points it passes through.
- plug this and either point into the slope-point equation: y - y1 = m(x - x1)


Cartesian plane and graphing 3
The only quadrant which contains no points of the graph of y = -x2 - 8x - 14 is quadrant

a. I
b. II
c. III
d. IV
e. There are points on this graph in all four quadrants.
Why not complete the square to put the quadratic into the form \(\displaystyle \mbox{y = -(x - k)^2 + h}\) and sketch.
 
josh123 said:
Hey guys I need some help. Can someone show me how to do these problems?


Line thru 2 pointsAn equation for a line passing thru (1, 2) and (4, 6) is

a. 4x - 3y + 2 = 0
b. 4x + 3y + 2 = 0
c. 4x - 3y - 2 = 0
d. 4x + 3y - 2 = 0
e. none of these


Cartesian plane and graphing 3
The only quadrant which contains no points of the graph of y = -x^2 - 8x - 14 is quadrant

a. I
b. II
c. III
d. IV
e. There are points on this graph in all four quadrants.


Thanks,
Josh

Not sure if I am right but worth a try
 
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