Explain how you determine an equation for a line using a table of input/output values.

1 answer:

6 0

Example :

x y
1 3
2 6
3 9
4 12

first thing u do is pick any 2 points (x,y) from ur table

(1,3) and (2,6)
now we sub those into the slope formula (y2 - y1) / (x2 - x1) to find the slope
(y2 - y1) / (x2 - x1)
(1,3)....x1 = 1 and y1 = 3
(2,6)...x2 = 2 and y2 = 6
sub
slope = (6 - 3) / (2 - 1) = 3/1 = 3

now we use slope intercept formula y = mx + b
y = mx + b
slope(m) = 3
use any point off ur table...(1,3)...x = 1 and y = 3
now we sub and find b, the y int
3 = 3(1) + b
3 = 3 + b
3 - 3 = b
0 = b

so ur equation is : y = 3x + 0....which can be written as y = 3x...and if u sub any of ur points into this equation, they should make the equation true....if they dont, then it is not correct

and if u need it in standard form..
y = 3x
-3x + y = 0
3x - y = 0 ...this is standard form

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Answer:

$(y-7)^2+(x-2)^2=16$

and

$(x+2)^2+(y-15)^2 = 9$

Step-by-step explanation:

The standard equation of a circle is $(x-h)^2+(y-k)^2=r^2$ where the coordinate (h,k) is the center of the circle.

Second Problem:

We can start with the second problem which uses this info very easily.
(h,k) in this problem is (-2,15) simply plug these into the equation. $(x--2)^2+(y-15)^2=r^2$ .
We can also add the radius 3 and square it so it becomes 9. The equation.
This simplifies to $(x+2)^2+(y-15)^2 = 9$ .

First Problem:

The first problem takes a different approach it is not in standard form. But we can convert it to standard form by completing the square.
$y^2-14y+x^2-4x+37=0$ first subtract 37 from both sides so the equation is now $y^2-14y+x^2-4x=-37$ .
$y^2-14y+x^2-4x+37=0$ by adding $(-\frac{b}{2a} )^2$ to both the x and y portions of this equation you can complete the squares. $(-\frac{b}{2a})^2=(-\frac{-14}{2(1)})^2$ and $(-\frac{-4}{2(1)})^2$ which equals 49 and 4.
Add 49 and 4 to both sides and the equation is now: $y^2-14y+49+x^2-4x+4=-37+49+4$ You can simplify the y and x portions of the equations into the perfect squares or factored form $(y-7)^2$ and $(x-2)^2$ .
Finally put the whole thing together. $(y-7)^2+(x-2)^2=16$ .