How do you graph Quadratic Systems?

Question

The ones in the book to go by are:
x^2+y^2=25 which is a circle
y-x=1 which is a line
Where they intersect are the two solutions. I can't figure out how they graphed these in the book and none are in earlier chapters. Also couldn't find any websites that explain how.
How do you graph them?

Thanks darkmagician you really solved my problem! Somebody; anybody else please!!

Answer 1

x^2+y^2=25 is a circle with radius 5 with center on the origin. y-x=1 is quivalent to y=1+x which is a line with slope of 1 and intersects the y axis at 1. You could graph them and see that the intersection points are (4,5) and (-3,-2) but solving them analytically is safer. To check the observation that the intersection is indeed (4,5) and (-3,-2), you solve for one variable in one equation and inject it into the other equation. Since we know that y=1+x, we substitute y with 1+x in this equation: x^2+y^2=25.

x^2+(1+x)^2=25
now it becomes 2(x^2 +x -12)=0
finding the zeroes 2(x-4)(x+3)=0
we see from here that x-4=0 and x+3=0 and therefore x=4 and x=-3 are equally correct.

Substituting these two values in y-x=1 or x^2+y^2=25 will give us the corresponding y value for each x value respectfully.

Therefore, the intersection points are (4,5) and (-3,-2)

Answer 2

4 squared = 16. and 5 squared = 25 and 16 + 25 = 41 which is not 25, so the above answer is wrong. In addition, -3 squared = 9. and -2 squared = 4 and 9 + 4 = 13.
However, the description of the circular graph is correct.

First, a negative number squared is a positive number. (-5 * -5) = 25 So, your number can run from 0 to 5 for both x and y. In each quadrant you will have x being the square root of 0 to 25, while y is the square root of 25 to 0. (sq root X = 17 plus sq root y = 8) (They are reciprocals.) So, the numbers will form an arc in each quadrant.

Obviously, y-x=1 is an easy line to draw, (y = - 3, x = - 4) with subtracting a negative number = a positive number.

I will not solve the problem for you, but I hope it is fairly obvious how to do it now.


Dear Doug: First, I got the right answer, you did not. Second, Get a life -- need to insult everyone without provocation because of your inadequacy? As I said, whodroppedth correctly described how to draw the circle, except for your brilliant insight of using a compass. So, we had a circle and a line. The answers are (X then Y) +3 and +4 and -4 and -3, which you should have been able to figure out if you understood what you were talking about. No slop if you knew the answer. I assume brainmonkeey passed fifth grade and can draw both a circle and a line, but obviously you find it a challenge, since you spent so much time trying to explain it. Since this is a math problem perhaps the purpose was to understand what the circle meant. Minor detail understanding rather than drawing blindly. Also, the exercise was to work from a graph (duh) not work out the answers as an equation. So, I'll go back on my meds, and you can go back to fifth grade, where among other things you can learn to multiply - at least, up to 5 times 5, so next time you, too, can get the right answer.

Answer 3

Rehabob needs to get back on his meds

The equation for *any* circle with its center at (h,k) and a radius of r is simply:

(x-h)^2 + (y-k)^2 = r^2

In your problem, h = k = 0 so life is much simpler. Rearrange the equation for the line to get

x = y-1

then substitute y-1 for x in the equation for the circle.

(y-1)^2+y^2 = 25

or

2y^2-2y+1 = 25

and then

2y^2-2y-24 = 0

and, factoring out the 2, you get

y^2-y-12 = 0

Solve the quadratic using the quadratic formula. (Actually, this one is easy enough that you should be able to factor it ) If there are two real answers, the line intersects the circle at two points. Back substitute the two values of y into x = y-1 to find the x values that go with them.

If there had only one real answer, it means the line is tangent to the circle. Do the same thing to get the x value that goes with it.

If there are no real roots, it means that the line doesn't intersect the circle.

In this case there will be two real values for y and I'll let you do the arithmetic. (I think 'whodroppedth' got it right, but double check )

To graph the circle, get a compass and set it to 5 unit marks on your graph paper (the radius of the circle from the given equation is 5 units) Put the point of the compass at the origin (0,0) and draw your circle.

To graph the line, go to -6 on the x-axis (6 units to the left) and then straight down to -5 units in the y direction. Make a dot. Then to to +6 units on the x-axis and go up to +7 units in the y direction. Make another dot. Now use a straight edge (like a ruler) to draw a line between the two dots.

Where your line and circle intersect should be the same coordinates as you calculated (making allowance for a bit of 'slop' in the drawing )


Doug

Answer 4

use graph paper