Ask a few questions about the second function and ask a question about the second function

1) Is the point (-1,2) in your graph on a proportional function? If you are, let the analytic formula of the proportional function be y=kx, because the point is crossed (-1,2), so k=-2, so the proportional function is y=-2x, and the primary function obtained after translating one unit to the left is parallel to the proportional function, you can let the primary function be y-2x+b, and the primary function crosses the point (-1,0), so 0=2+b, and b=-2, so the analytical formula of the primary function is y-2x-2

2) The intersection of two function images at one point on the x-axis means that the intersection point of the two functions and the x-axis is the same, and the intersection point of the function y=x+1 and the x-axis is.

1,0), so the intersection of y=nx-2 and the x-axis is also (-1,0), so 0=-n-2, and the solution is n=-2

3) Since the primary function has no intersection with y=2x-3, the primary function is parallel to y=2x-3, so the analytical formula of the primary function can be set to y=2x+b, and b is not equal to -3 (because if it is equal to -3, the images of the two functions coincide). Since the primary function intersects with y=6-x at the point (5,k), it means that the point is on y=6-x, so k=6-5=1, so the point (5,1) is solved on the required primary function image, i.e., 1=2*5+b, and the solution is b=-9, so the analytical formula of the primary function is y=2x-9

2.That intersection is on the x-axis, so the ordinate is 0, i.e., (m,0), (m,0) is on y=x+1, substituting m=-1, (-1,0) substituting y=nx—2, and getting n=-2

k) on y=6-x, substitution to obtain k=1,a(5,1).There is no intersection between the image of the primary function and the straight line y=2x-3, that is, the two straight lines are parallel and k is equal, let the primary function be y=2x+b, and the point a is substituted to obtain b=-9, then the analytical formula of the primary function is y=2x-9

y=-3/4x+b

x=0,y=b

y=0,x=4b/3

then ob=b, oa=4b 3

By the Pythagorean theorem.

ab= [b +(4b 3) ]=5b 3, so the perimeter b + 4b 3 + 5b 3 = 16

4b = 16b = 4, so area = ob*oa 2 = 2b 3 = 32 3

First of all, k=2 is obtained from the parallel relation, and y=2x+b is obtained, and m(0 4) is substituted to get b=2, so b is chosen

If the parallel slope of the two lines is equal, then k=2, bring in (y=2x+b, b=4, so choose b

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Only the concept of junior high school and the calculation of elementary school are needed!!

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Substituting the origin coordinates (0,0) into the equation, 0=m+1, m=-12m-2>0, that is, m>1.

The intersection point with the y-axis is the intersection point of x=0, and y=m+1>0, that is, m>-1.

2m-2<0 and m+1>0,-10,y=10 when x=0, y=30 when x=10

10=b 30=10k+b k=2

The function is y=2x+10

If k < 0, x = 0 when y = 30, x = 10 liquid code y = 1030 = b 10 = 10 k + b k = -2

The function is y=-2x+30

3.The circumference of the isosceles triangle is 40cm

y=40-2x.

0

;The condition for the image to pass the origin is that when x = 0, y = 0, and the substitution function gives itself an equation.

The necessary condition for the increase of the requirement is that the coefficient of the function x is a positive number, that is, the coefficient is greater than zero.

The meaning of the intersection above is: when x=0 y=m+1, the value is greater than zero, and the meaning of these three quadrants is that the intersection with the y-axis is above the x-axis, and as x increases, y decreases, that is, the coefficient of the function x is less than zero, and the two conditions are combined.

1(1) over the origin (0,0) 0=(2m-2) 0+m+1 m=-1

2) Y increases with x by 2m-2 0 m 1

3) The intersection of the function Wu Langshen image and the y-axis is m+1 0 m -1 above the x-axis

2 Let y=kx+b k≠0

If k 0 0 x 10 0 kx 10k b kx+b 10k+b

Libu 10 y 30 b=10 10k+b=30 cavity k=2 b=10 y=2x+10

If k 0 0 x 10 1 0k kx 0 10k+b kx+b b

10≤y≤30 ∴ b=30 10k+b=10 ∴k=-2 b=30 ∴ y=-2x+30

3(1)y=40-2x

2) 0＜40-2x＜2x

10＜x＜20

Solution: y = kx-2 and the intersection points of x,y axes are p(2 k,0),q(0,-2), respectively

Simultaneous y = k x Yes.

k/x = kx-2

kx²-2x-k = 0

x = 1+ (1+k )]k (negative rounded) then y = k x = k [1+ (1+k )]r([1+ (1+k )]k,k [1+ (1+k )]s opq = s prm

1 2·op·oq = 1 2·pm·rm2 k·2 = k [1+ (1+k)] This equation looks a little complicated, in fact, it is not difficult, do not blindly multiply the two ends of the equal sign by the algebraic elimination denominator, in that case, it is easy to become a higher-order equation, I write out the process for the landlord's reference.

4/k = k - 2k/[1+√(1+k²)]4/k = k(1-2/[1+√(1+k²)]4 = k²·[1+k²)-1]/[1+√(1+k²)]4 = k²·[1+k²)-1]²/k²

4 = 1+k²)-1]²

2 = 1+k²)-1

k = 2√2

Translate 2 units to the right, and then translate 3 units upwards, according to the analytic formula of "x left plus right subtraction, y plus plus subtraction".

y=3(x-2)+1+3, simplified to y=3x-2

Then choose A.

Let x=0, then y=0+1=1

So y=3x+1 over (0,1).

Shift it to the right by 2 unit lengths, and then translate it upwards by 3 unit lengths is (0+2,1+3), that is, after the translation (2,4), the x-coefficient remains unchanged.

So y=3x+b

(2, 4).

4=6+b,b=-2

So it's y=3x-2

Substituting the origin coordinates (0,0) into the equation, 0=m+1, m=-12m-2>0, that is, m>1.

The intersection point with the y-axis is the intersection point at x=0, and y=m+1>0, which is m>-1.

2m-2<0 and m+1>0,-10,y=10 when x=0, y=30 when x=10

10=b30=10k+b

The k=2 function is y=2x+10

If k < 0, x = 0 y = 30, x = 10 when y = 1030 = b10 = 10k + b

The k=-2 function is y=-2x+30

3.The circumference of the isosceles triangle is 40cm

y=40-2x.

00<40-2x<40

A straight line that connects points (0,40) and points (20,0) with 0.

(1) The function must first be an equation, so the first question is not proportional to the value of k.

2) x+2y=20, the analytical formula of the function (y=, the value range of x is 0