Junior 2 Mathematics questions about inverse proportional functions

y1 is proportional to x, let y1=mx

y2 is inversely proportional to x-2, let y2=n (x-2) substitute the above two equations into y3 y1 3 y2 to get y=3mx-3n (x-2).

When x 1 is added again, y=-1. When x=3 and y=13 are substituted into the above equation respectively, we get 1=3m 1-3n (1-2) (1).

13=3m×3-3n/(3-2) (2)

Solve the equation composed of equations (1) and (2) to obtain.

m=1 n=-4/3

Substituting the values of m and n into the initial equation gives the analytic formula y=3x+4 (x-2).

y1 is proportional to x, let y1=mx, y2 be inversely proportional to x-2, let y2=n (x-2).

Rule. y=3mx-3n/(x-2)

Substituting x=1, y=-1 gives -1=3m+3n

Substituting x=3, y=13 gives 13=9m-3n

The simultaneous solution yields m=1, n=-4 3

then y=3x+4 (x-2).

Let y1=k1x, y2=k2 x-2

then y=3*k1x-3*k2 x-2

Substitute x 1 and y = -1.

1=3k1-3k2/-1

1=3k1+3k2

k1+k2=-1/3

Substitute x=3, y=13.

13=9k1-3k2

13/3=3k1-k2

So k1+k2=-1 3

3k1-k2=13/3

So 4k1=4

k1=2, so k2=-7 3

So y=6x+(7 x-2).

Let a(a,b) then b=1 a ab=1

The ordinate of point b is - b substituted by :ap=bh and the function formula: -b=1 x x=- 1 b=- a b(-a,-b).

So: s triangle aph = (1 2) ph * pa = (1 2) * (2a) b = ab = 1

S triangle bph = (1 2) ph * pb = (1 2) * (2a) b = ab = 1

Therefore, the AHPP area of the quadrilateral is: 1+1=2 (is a fixed value).

a, b are any two points on the image of the function y=1 x with respect to the origin o symmetry, so a(x,1 x)b(-x,-1 x).

bh|=|ap|

ap is parallel to the y-axis, the x-axis is at the point p, the bh is parallel to the y-axis, and the x-axis is at the point h, so ap bh

So ahbp is a parallelogram.

Area = |ap|*|ph|=x*2y=2x*1 x=2 is a fixed value.

I wrote it with a formulator and can't show it here.

1. y=600 15x (x is greater than or equal to 1, less than or equal to 600).

2. Let the function of y and x be y=k (, and substitute x=,y= into k=, so the functional relationship between y and local excitation x is y= (,

y=3/|x|

When x>0 y>0

When x>0 y>0

Image in section. 1. 2 quadrants.

After (-1,3).

Third Quadrant. Because x, no matter what value is taken, xy=3, it is in one or three quadrants; But because x=-1, it's in the third quadrant.

Because the point through which the image passes can be pushed, when x=-1, y=3, the image must pass the second quadrant, and because the function is proportional, it can only be passed.

2. Four quadrants.

From the known conditions, y1=k1 x, y2=k2x substitute x=-1 to get y1=-k1, y2=k2, so y1+y2=y=0=k2-k1=0

Of course, k1-k2 is also 0So choose B.

by=y1+y2 ……

y1=k1/x ……

y2=k2*x²……

Synopid: y=k1 x+k2*x ......Substituting x=-1 and y=0 yields -k1+k2=0

It is known that y1=k1 x, y2=k2x

y=k1/x+k2x²

And because when x = -1, y = 0, so -k1 + k2 = 0, that is, k1-k2 = 0

The answer is bIf you don't understand, you can ask me again.

y=k1 x+k2*x 2, bring x=-1, y=0 into the knowable relationship and select b

(1) Bring a set of values in and find out.

Analytic formula for proportional function: y=[-(root number2) 2] x inverse proportional function analytical formula: y=(-2 root number2) x

2) Since x=a, y=negative root number 2 is a set of corresponding values of the proportional function, then it is easy to find x=a=-2 by substituting the analytic formula

Because y=(-2 root number 2) x

So substitute x=-2 and y=negative root number 2.

Negative root number 2 = (-2 root number 2) -2

Obviously, the above formula does not hold.

So this set of values is not the corresponding value of the inverse proportional function.