Questions about the first function!

1.Solution: Obtained from the question.

2=（-2）*a-3

It can be solved a=2Solution: Obtained from the question.

a-1<0, so a<1 is solved

3.Solution: Obtained from the question.

Simultaneous two-line equations.

y=x+1y=-x+1

It can be solved with x=0 y=1

4.Solution: Obtained from the question.

1) When the slope of the line is 0, a= intercept is, disagree, rounded.

2) When the slope is not 0, then 2a-3>0 is obtained, and the intercept of 2-a > or = 0 is solved to get a> or =2

5.Solution: Obtained from the question.

First, we get the equation for the straight line.

Let the equation for the line be y=kx+b

By substituting the coordinates of the two points, k=-2 3 b=1 3 can be solved, that is, the linear equation is y=-2 3x+1 3

The obtained longitudinal intercept is 1 3 and the cross section is 1 2

Therefore, the area of the triangle is 1 3*1 2 2=1 12

The point is on a straight line, and the coordinates of the point can be directly substituted into the linear equation.

2. a<1

y decreases with the increase of x, and the function is a monotonic subtraction function, y=kx+b, and the slope k is less than 0

The intersection of two straight lines must be on two straight lines, i.e. y=x+1=y=-x+1

4.3 2 The straight line is not the second quadrant, that is, y = kx + b, the slope k (to determine the direction of the straight line) is greater than 0 and b (the intersection point with the y axis) is less than 0

For straight lines that pass through the points (-1,1) and (2,-1), the coordinates of the two points can be substituted into the system of equations of y=kx+b, 1=-1k+b, and -1=2k+b

We get the equation k=2 3, b=1 3, and the linear equation y=2 3x+1 3

The distance from the intersection of the line and the coordinate axis to the coordinate axis is the right-angled side of the triangle enclosed by the line and the two coordinate axes.

Substituting (0,y)(x,0) into y=2 3x+1 3 respectively yields (0,1)(-1 2,0).

The area is the bottom (|-1/2|) multiplied by height (|1|Divide by 2

Area s=1 2|1|*|1/2|=1*1/2*1/2=1/4

:y=-2 substitution.

2.<1 ：a-1<0

3.(0,1): Simultaneous solution.

4.>2: 2a-3>0,2-a<0 simultaneous solution:

The slope [1-(-1)] [(-1)-2]=(1-y) (-1-0)=(1-0) (-1-x) is obtained to obtain x=1 2, y=1 3 is the side length of the triangle.

Get: y=(4 3)x

2。From the parallel, it can be known: k-pin type.

Then substitute the bucket segment point (0,3) to obtain: b

b>0,b<0

k>0, b<0 passes.

Three, four, one quadrant, k<0, b>0 passes.

2, 1, 4 quadrants, k<0, b<0 passes.

Two, three, and four quadrants.

I hope it can be helpful to the landlord, and at the end I warn the landlord: I don't deny that you can have a great future without reading, but I want to say that if you are reading the reputation again, at least you must understand the most basic knowledge. Of course, this is just a personal opinion.

The number of copies of this kind of evening newspaper purchased every day for a month.

According to the title, these newspapers are sold out every day.

So the profit of the month: 100 * (yuan.)

Buy the first number of the good ears of the newspaper every day for a month.

Newspaper sold 20*150*

Returned newspaper 10*30*(

Sold - returned = socks Sakura = 420-30 = 390 yuan.

2）y=20x*

2x+120-x+120

x+240 can be seen from the relation, the greater the x, the greater the profit.

However, 120 is less than or equal to x less than or equal to 200

Therefore, the maximum profit value of the month = 200 + 240 = 440 yuan.

y=x* at 0 x 100

x 100 y=65+(x-100)*(24 30) Question 2 62 degrees Bringing in method 1 Electricity charges payable.

2. Bring in the equation to get x=50

a functional relation to x; Let : (y 2) (2x 1)=k, when x -1 y 2 is substituted into k = -4, y = -8x-6

2.Point m(a,-2) on this function image, 2=-8a-6, a=-1 2

3.If -3 x 0, the value range of the file start group y; 18≤y≤-6.

4.If 2 y 10, the value range of x is: x=(-y-6) 8 , 2 x -1

2) If the flow velocity of the water is a meter (quantitative), then the functional relationship between the water inflow q (cubic meter) per branch orange clock and the diameter d (m) of the selected water pipe is: q = 15 atd 2 - where the independent variable is t (minute) - the constant is 15 ad 2——.

3) Knowing the primary function y x 6-m, find:

, the function image intersects the y-axis on the positive semi-axis.

, the image passes through the origin.

Answer: The straight line passes through the origin: y=kx

Passing point (1,3): k*1=3, k=3

So: y=3x

p on the x-axis.

then y=0y=3x+6=0

x=-2, so p(-2,0).

So y=kx+b.

2=k+b0=-2k+b

k=2/3，b=4/3

y=2x/3+4/3

When y=0: 3x+6=0

3x=-6x=-2

So the point p coordinate is (-2,0).

Put (1,..)2) (-2,0) substitution: k+b=2 -2k+b=0

Yes - get 3k=2

k=2 3 substitut k=2 3 into b=4 3

So y=2 3x=4 3

The straight line y=3x+6 passes the point p, and the point p is on the x-axis, y=0, i.e., 3x+6=0, x=-2 point p(-2,0).

So y=kx+b over (1,2)(-2,0).

i.e. 2=k+b

0=-2k+b k=2／3 b=4／3

Put the dot (1,..)2) Bring in y=kx+b

and intersect with the x-axis at the point p

So p is (x,0).

Bring p to (x,0) to y=3x+6

0=3x+6

3x=-6x=-2

So p is (-2,0).

Bring p to (-2,0) to y=kx+b

0=-2k+b

So 2=k+b

0=-2k+b ②

Solve the system of equations.

Get. 2=3k

k=2 3 brings k=2 3 in

2=k+b2=2/3+b

b=2-2/3

b=4 3, so y=kx+b is analytically y=2 3x+4 3, and to do this kind of problem, we need to grasp the properties of the function and the meaning of "and intersect the point p with the x-axis".

The line y=3x+6 intersects the x-axis at the point (, then.

The image of the primary function y=kx+b passes through the point (1,..)2) （

k=2/3 b=4/3

The line y=3x+6 intersects the x-axis at the point p

When y=0, x=-2

So the point p coordinate is (-2,0).

Because once the function y=kx+b of the image passes through the point (1,.2), and the intersection with the x-axis at the point p is obtained by the two-point formula:

y-2)/(x-1)=(y-0)/(x+2)(x+2)(y-2)=y(x-1)

xy-2x+2y-4=xy-y

3y=2x+4

y=2x/3+4/3

y=kx+b image crossing point (1,..)2)，2 = k +b, b = 2-k

and intersects the x-axis at the point p, the coordinates of p are y=0, x= -b k = (k-2) k

The straight line y=3x+6 also passes through the point p, 0 = 3(k-2) k +6, k = 2 3, b = 4 3

The analytic formula for this primary function is y = 2 3x + 4 3

The image is crossed by the primary function y=kx+b (1,..)2) Know 2=k+b

and intersects with the x-axis at the point p, if the straight line y=3x+6 also crosses the x-axis through the point p.

y=03x+6=0

x=-2 is (-2,0).

Bring in y=kx+b

is -2k+b=0

Put Lianli.

Solution k=2/3

b = 4/3

So the analytic formula is.

y = (2/3) x + 4/3

From the known: y=kx+b passes through the point (1,2), then k+b=2, and intersects with the x-axis with the point p, so that y=0, then x=-k b, that is, p(-k b,0), and brings in the second straight line equation: -3k b+6=0, simultaneous k+b=2, k=2 3, b=4 3, so the analytical formula is y=2x 3+4 3

1. The origin is a one-time function, and the constant term is 0

m-2=0m=22、

If y decreases with the increase of x, the x coefficient is less than 0

8-2m<0

m>43, image passing.

One, two, and three quadrants.

Then the x-factor is greater than 0, and the intercept in the y-axis is greater than 0

8-2m>0,m<4

m-2>0,m>2

So 2

(1) Because it passes through the dot, it is a proportional function, so m-2=0, so m=2

2) Because it decreases as x increases, so k=8-2m0 so m4(3) because of the elapse.

One, two, three quadrants So y increases with the increase of x So k = 8-2m 0, so m 4 because it passes through 2 quadrants so m-2≠0 so m≠2 so m 4 and not equal to 2

The straight line l:y=mx+5m,a(-5,0),b(0,5m), 5m=5,m=1 obtained by oa=ob, and the analytical formula of the straight line is: y=x+5

AM is perpendicular to OQ, Bn is perpendicular to Oq, so angle amo = angle BNQ = 9o° Bn parallel AM (isotope angles are equal, two straight lines are parallel).

Angle ABN = Angle BAM = 180° (two straight lines are parallel, complementary to the inner angles of the same side) and angle Bao + Angle ABO=9O° (mutual remainance).

Angle mao + angle obn = 90°

Angle mao + angle AOM = 90°

Angular AOM = Angular obn

aom≌△bon

Finally, we get bn=3

After E as an extension line perpendicular to the OP, it can be proved that EMB is fully equal to AOB, (as for how to prove, please think for yourself) so EM=OB, and OB=BF, EM=BF, and EM is parallel to BF, EMP is full equal to OBF, MP=BP, so that the outer y=0, x=-5, ao=me=5, PB=MP=5 2= is a fixed value.

By making the extension line of EM perpendicular to OP, it can be proved that EMB=OB, and OB=BF, so EM=BF, and EM is parallel to BF, so EMP is all equal to OBF, MP=BP, so that Y=0, X=-5, so AO=ME=5, PB=MP=5 2= is a fixed value.