Junior high school math problems are in a hurry, please take steps

Solution: According to the known conditions and the vertex coordinates of the parabola, the following three formulas can be obtained.

a-b+c=0

b/2a=1

4ac-b^2)/(4a)=-4

solution, a=1

b=-2c=-3

The analytic formula is y=x 2-2x-3

x2 = 3b point coordinates (3,0).

The coordinates of point c are (0,-3).

Let the coordinates of the q point be (x,y), then.

qc^2=x^2+(y+3)^2

qb^2=(x-3)^2+y^2

qc=qbx^2+(y+3)^2=(x-3)^2+y^2y=x^2-2x-3

The solution is that x1=(1+ 13) 2, x2=(1- 13) 2y1=-(1+ 13) 2, y2=-(1- 13) 2q point coordinates (x1, y1) or (x2, y2) x1, y1, x2, y2 data is too complicated, you can substitute it yourself.

y=ax^2+bx+c

Because x2=1, y=0

So a+b+c=0

then c=-a-b

Because a>b>c, 3a>a+b+c=0

So a>0

It can be obtained by a>b>c and c=-a-b.

a>b>-a-b

Divide the two sides by a.

1>b a>-1-b a, ie.

b/a<1

b a>-1-b a, i.e. b a>-1 2 so. 1 2 Because according to the Vedic theorem.

x1+x2=-b/a

So. 1<-b a<1 2, ie.

1<(x1+x2)<1/2

Because x2 = 1, it is -2

y=ax 2+bx+c intersects the x-axis at the points a(x1,0),b(x2,0)(x2>x1), x1,x2 are the two roots of the equation x-square-2(m-1)x+msquare-7=0, and x1 square + x2 square = 10

x 2-2 (m-1) x + m 2-7 = 0, according to Veda's theorem.

x1+x2=2(m-1)

x1*x2=m^2-7

x1+x2)^2=x1^2+x2^2+2x1*x2=10+2(m^2-7)

4(m-1)^2=10+2(m^2-7)

The solution is m=2 and the equation is x 2-2x-3 = 0

Substituting the Vedic equation gives x1=-1, x2=3(x2>x1) because ax 2+bx+c=0 is the same root as x 2-2x-3=0.

So y=f(x)=ax 2+bx+c=k(x 2-2x-3) (k is a non-0 real number).

Then according to the ordinate of the vertex m is -4, the abscissa is -b, the ab coordinate is (1, -4), the axials of 2ag(x)= x 2-2x-3, so k=1, that is, the parabolic function, the analytic formula is: y=f(x)=x 2-2x-3, so the coordinates of ab point are (-1,0) and (3,0).

Intersect with the y-axis at the point C, the coordinate of the point C is (0,c), i.e. (0,-3) I am yj (3) It's too troublesome.

Solution: (1) Let this parabola be.

y=a（x+1）^2

from known conditions.

a（1+1）^2=-4

solution, a=1

The analytic formula for this parabola is .

y=-（x+1）^2

Because AD is the angular bisector of ABC, so.

bad=∠cad

And de ac, df ab

So 1= cad

2=∠bad

i.e. 1= 2 equal.

∠1=∠2

ad is the angular bisector, then bad= cad

and bad= 2 cad= 1 (wrong angle within parallel lines) 1= 2

The quadrilateral AEDF is prismatic.

∠1=∠2

Because de is parallel to ac

So 1= dac

Because DF is parallel to AB

So 2= bad

Because AD is the angular bisector of the BAC.

So dac= bad

So 1= 2

ad bisects the angle bac, so the angle bad=angle cad, df parallel ab, de parallel ac, so afde is parallelogram, so, angle bad=angle 2, angle cad=angle 1, so angle 1 = angle 2

∠1=∠2=1/2∠bac

Because AD is the angular bisector of ABC.

So bad= cad

And because of de ac, df ab

So 1= cad, 2= bad (two parallel lines with equal wrong angles) so 1= 2=1 2 bac

In addition, it can also be obtained by proving aed congruent AFD (corner edge), which seems to be a simpler method above.

These two angles are equal. The reason is: from the angle bisector of the angle of the angle EAF AD can be seen, the angle EAD = angle FAD, because the line segment DE and DF are parallel to AF and AE respectively, according to the theorem of equality of the inner wrong angle, the above conclusion can be drawn.

Equal relationship ha! DF parallel AB angle 2 is equal to angle EAD

deparallel AC angle 1 is equal to angle fad

ad bisects the angle bac so the angle ead equals the angle fad so angle 1 equals angle 2

Hope to give a good review Thank you.

1 = 2, prove: because de ac, then bed= bac, the same way dfc = bac, we know bed= dfc, because bed and aed sum is 180 degrees, dfc and afd sum is 180 degrees.

So aed = afd, and because ad is the angle bisector, the sum of the inner angles of the triangle is 180 degrees, 1 = 2

Because ad is an angular bisector so the angle bad=angular dac because de is parallel to ac so <1=< DAC is the same as DF parallel ab, so "2="bad< so1=<2< p>

∠1=∠2

de ac, dea and bac are complementary angles to each other, df ab, dfa and bac are complementary angles to each other, dea = dfa

and ad is the angular bisector of abc, bad= cad

AFDE is a parallelogram.

Because angle 2 = angle bad angle 1 = angle cad angle bad = angle cad

So angle 2 = angle 1

The quadrilateral afde is a parallelogram.

AD is the angular bisector of the triangle ABC, so 1=2

Solution: Equal.

Rationale: Because ED is parallel to AC and DF is parallel to AB

So angle 1 is equal to angle daf, angle 2 is equal to angle ead

And because AD bisects the angle BAC

So the angle EAD is equal to the angle FAD

So angle 1 is equal to angle 2

equal, according to the nature of parallel lines.

（1）.

2x^2/3-x/6-1/2=0

4x^2-x-3=0

4x+3)(x-1)=0

The solution is: x1=-3 4, x2=1

2）x(2x-7)=-49/8

2x^2-7x=-49/8

2x^2-7x+49/8=0

2(x-7/4)^2=0

Solution: x=7 4

3). x+1)(x-1)=2√2x

x^2-2√2x-1=0

x^2-2√2x+2=3

x-√2)^2=3

The solution is: x1 = 2 + 3 , x2 = 2- 3 (4).2x^2-10x=3

x^2-5x+25/4)=31/4

x-5/2)^2=√31/2

Solution: x1=(5+ 31) 2 , x2=(5- 31) 2

This topic is relatively simple, learn to use the Internet, but don't rely on the Internet, then you won't improve!

Suppose y1 = kx and y2 = m x

So when x=1, y1=k, y2=m, y=k-m=3, when x=-2, y1=-2k, y2=-m 2, y=2k+m 2=15 2

The two formulas can be solved.

k=m=so the solution is y=

Be is the angular bisector so the angle abe = angle ebc because it is a parallelogram so the angle aeb = angle ebc so the angle abe = angle aeb so the triangle abe is an isosceles triangle so ab=ae=2 the parallelogram is equal on opposite sides cd=ab=2 bc=ad=ae+ed=3 so the perimeter = 10 take it.

Because be bisects the angle abc, ab is equal to ae and is equal to 2Because it's a parallelogram. So dc is equal to ab is equal to 2, ad is equal to bc is equal to ae plus de is equal to 1+2=3, so the perimeter is (3+2) multiplied by 2 is equal to 10

In the parallelogram ABCD, ad bc, ab=cd, ad=bc, ad=ae+ed=2+1=3

And because be is the angle bisector of the angle abc, the angle abe = angle ebc because ad bc, so the angle aeb = angle ebc

So the angle abe = the angle aeb

So ab=ae=2

So perimeter = ad+dc+cb+ab=10

Because of AD BC

So the angle AEB is equal to the angle EBC

and be is the bisector of the angle abc.

So the angle ABE is equal to the angle EBC

So the angle AEB is equal to the angle ABE

So the triangle ABE is an isosceles triangle.

So AE is equal to AB

So the circumference is 2+1+2+2+3=10

Be is the bisector of ABC.

abe=∠cbe

ad∥bc∠aeb=∠cbe

abe=∠aeb

ab=ae=2 cd=ab=2

and bc=ad=ae+de=2+1=3

The circumference of the parallelogram ABCD is AB+CD+AD+BC=2+2+3+3+10

Because ad bisects the angle abc, the angle abe = angle ebc because it is a parallelogram, so ad bc, so the angle aeb = angle ebc so the angle abe = the angle aeb

So ab=cd=ae=2

So perimeter = 3 + 2 + 3 + 2 = 10

Parallel, again the angle bisector, so ae=ab

Because be bisects abc, abe= ebc and ad bc, aeb= ebc

So ab=ae

So the circumference is 10

Solution: The analytic formula of this function y=kx+b

x=-4,y=9;x=6, y=3.

9=-4k+b

3=6k+b

k=b=y=function f(x) is an even function defined on r, so, f(x)=f(-x), when x is greater than or equal to 0, f(x)=-x 2+4x,, let, x<0, then -x>0, so, f(-x)=-x 2-4x,,,f(x)=f(-x)f(x)=-x 2-4x,f(x).

The analytic formula can be written in two cases: x<0 and x>=0.

x 2-mlnx-x 2+x=x-mlnx 0(x>1),x mlnx,m x lnx,let g(x)=x lnx,g'(x)=(lnx-x*1/x)/(lnx)^2

lnx-1) (lnx) 2, take g'(x)=0, the solution gives lnx=1, x=e, because g(x) decreases monotonically on x (1,e) and increases monotonically on x (e,+, so the minimum value is obtained at x=e, gmin(x)=g(e)=e, so there is m e;