How do I graph 4x+y=0 - Office Equipment & Supplies

The easiest way to solve and graph and equation is to put the equation into the slope intercept form y = mx + b, where m is the slope and b in the y-intercept.

To do this, we subtract 4x from both sides and get y = -4x + 0.

From this we know m = -4 (slope) and the y-intercept is 0.

I always start with the y-intercept and put a point there. Thus, we have a point at (0,0). Using this as a starting point, we now use the slope of -4 to get future points. Since it is negative, we go one unit to the left and up four units. So we have the point (-1,4). Using a ruler, we connect these points and continue on both sides to produce the line.

There is also a great free online program/app called Desmos that you can use to check your work. Type in the equation of the line and it will graph it for you.

Good luck,

Paul

See image below

First, I graphed the lines and the point using Desmos.com.

I noticed that the two lines are perpendicular to each other and the point (1,-1) appears to be on the right side of the circle, on a line parallel to 3x -4y-10=0. The equation of this line is y= 3/4x - 1.75. The y-intercept is -1.75. Now we have two points on the opposite sides of the circle, (1, -1) and (0,-1.75). The midpoint formula will give you the centre of the circle and the distance formula will provide the radius.

Let me know if you have any questions.

Good luck.

Paul
Desmos Beautiful Free Math

You need to draw the curve y= x^2-4x+ 3. It is a parabola. The equality (=) is verified by all points on the curve. The curve divides the whole plane in two regions. In one region y will be less than x^2-4x+3. In the other regions y will be larger than x^2-4x+3
The regions are the interior of the parabola and its exterior. To find out the region where your inequality is verified you proceed as follow.

1. Look at the graph and choose the coordinates of one point. Select integer values (BTW the point (0,0) is usually a good candidate) and write y ? x^2-4x+3 where the ? stands for the inequality sign you will have to insert to make the relation correct.
2. Take x=0, and y=0 and substitute in y ? x^2-4x+3
3. 0 ? (0^2)-4*(0)+3.
4. Clearly the ? stands for < because 0 <3.
5. So at all the points in the region where (0,0) is located the condition y< x^2-4x+3 is TRUE.

In the other region y < x^2-4x+3, is FALSE and that second region (the one that does not contain the origin (0,0) is the one you need to find. It represents your solution set. Note that the origin (0,0) does not lie in the solution set region

You put x=0 then y=0 in turn.

The x intercept, when y=0, gives 4x^2-9 = 0 or x = 3/2 = 1.5

The y intercept, x=0, gives y = -9

First step is to find inequality in the form y<f(x), in this case y<5-0.5x. Then graphing it. See captured image

4x-(2x-3)=0
=>4x-2x+3=0 ; (after the bracket's oepned and multiplied by - [minus])
=>2x+3=0 ; taken x as common and 4-2=2 i.e. x(4-2) = 2x
=>2x=-3 ; +3's moved on the right-hand side, so it becomes -3
=>x=-3/2 ; 2 has been moved to the right-hand side and divided with -3 i.e 1/2 multiplied by -3 = -3/2

Hence, x = -3/2


Good luck.

Thanks for using FixYa.

Hii,you can use the following solution x^3 - 4x^2-12x=0 x(x^2-4x-12)=0 replacing -4x from -6x+2x x(x^2-6x+2x-12)=0
Now Factoring x(x(x-6) +2(x-6))=0 x(x-6)(x+2)=0 Solution is x=0, x-6=0,x+2=0 1) x=0 2) x-6=0 =>x=6 3) x+2=0 =>x=-2

use the slope-intercept form to graph y=2/3x+4

Hello,
Reset the memory to defaults and you will not get the error.
[2nd][MEM][7:Reset][Defaults] [2:Reset] [ENTER]. Press [CLEAR] and them [Y=] to enter the function. No need to set Zoom. Defaults work fine for y1= 4X+3.
Press [GRAPH].
Hope it helps