Two Quadratic Function Problems in Junior High School Finding Expert Process Solutions 10

13(1).The y values of a and b are zero, so that y = 0 in the function, we get x=6, x=-3, so ab=9

The value of x at point c is zero, so that in the function x=0, we get y=-9, so oc=9

2).First, find out that the area of the triangle abc is 1 2*ab*oc=

Because L is parallel to BC, so ed:bc=m:9, if you do the height on the side of BC, then the ratio of the height of the two is also m:9, so the ratio of the area of the two triangles is the square ratio of the two, and m squared: 81, so the S triangle ade=

The m range is 014(1).Let o be the origin, then the function can be written as, x=a*y 2, with the coordinates of point a (-8,4), then a=, so x= is the sought.

2).First find the coordinates of point b, because y=2, so x=-2, so b(-2,2), let p(x,0).

ap+bp = [(-8-x) 2+16] + [(-2-x) 2+4] under root

Since the minimum value of this equation is not easy to find, and the result must be greater than zero, it can be converted to find the minimum value of (-8-x) 2+16+(-2-x) 2+4.

After simplification, we get 2x 2+20x+88, that is, find the minimum value of x 2+10x+44=(x+5) 2+19, where -83).So op=5 is what is sought.

NOTE: 2 is the meaning of square.

1/2 of a A (-2 of a 3 a power) 3/4 of a A 5 of a 5.

The center point of the arch is used as the coordinate origin. The span of the arch is the x-axis and the arch height is the y-axis.

The coordinates of the three points are known from the question, respectively.

Left pier (,0), bridge vertex (0,, right pier (,0) let the quadratic function relation corresponding to the parabola be .

y=ax^2+b

b=a=So, the quadratic function relation corresponding to the parabola is. y=

Is it in the evaluation manual or in the textbook or in the supplementary exercises, I did it anyway, if you won't teach you tomorrow.

Answer: 1, from the known conditions: a-b+c=0, 9a+3b+c=0, (4ac-b

2)/4a=-2

a-b+c=0, 9a+3b+c=0, gives 4a=-b, a=-b 4

a-b+c=0, so c=5b 4

And because (4ac-b

2) 4a=-2, so (-5b.)

b2）/-b=-2

Push b=-8 9

So a=2 9, c=-10 9, and the expression is: y=2 9x -8 9x-10 9

2. The function and the y-axis intersect at the destruction point (0,1), so let the function formula be: y=ax +bx+1

The A in ax is mainly used to control the size of the opening of the parabola, so the shape of a parabola is known.

y=x, i.e., a=1

And because the axis of symmetry is a straight line.

x=-1 2, so Kiriwei b -2a=-1 2, so b=1

The resulting function formula: y=x +x+1

Solution: (1) From the meaning of the question:

y = x square + 20x (3 points).

The value of the independent variable x can be 0 x 25

2) y=- x squared + 20x

x-20) square + 200

a=-1 0,20 25, the maximum value exists.

When x = 20, y has a maximum of 200 square meters.

That is, when x=20, the area of the green belt that meets the conditions is up to ......I can't explain that.

Because the quadratic function image is symmetrical with respect to the straight line x=-2 where the vertex is located, according to the graph, the two intersections of the function image with respect to the x-axis are symmetrical with respect to the point (3,0), so the distance to the point (3,0) is 5 2The coordinates can be found,

It's too simple, the answer is the same as above.

Ke Xiangxiao finds the RT triangle abc, ac=8cmy=(ac*bc) 2-(1 2):*be*bf*sin b6x8 2-(1 2)*t*(10-2t)*(8 10)0 "shirt liter t<5).

From the known y=2x -4mx+m =2(x-m) -m, so that y=0 obtains: x=m+m*root number 2 2 or x=m-m*root number 2 2

So ab = absolute value (m * root number 2) so the area of abc s = 1 2 * absolute value (m * root number 2) * m = 4 root number 2

The solution yields m=2 or m=-2

m = plus or minus 2

The two intersection points are (1 plus minus 1 2 root number 2) m

The vertex ordinate is -m 2

So there is (plus or minus root number) 2m*m 2 2 = 4 times the root number 2 solution to m is plus or minus 2.

Forgive this disgusting process, but it can only be written in Chinese like this.

The original formula 2(x-m) m, so the coordinates of point c (m, m), xa+xb 2m, xa*xb m 2

So ab distance |xa－xb|(xa+xb) 4xa*xb 2m so area

So m 32 open to the 8th power.

ab=[root(-4m)2-8m2] 2

The ordinate of c = [8m2-(-4m)2] 4

sabc=ab*|yc|= 4 root number 2

Calculate it yourself later.。。

You're right to count high to 5.

The height is the ordinate, bringing y=5 into the function.

Get x = 4 or -2

So c(4,5) or (-2,5).

Thank you for adopting!

Hello! Here's the answer:

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

x=3,x=-1

So a(-1,0), b(3,0).

So ab=|-1-3|=4

That is, the bottom edge of the triangle is 4

Area = 1, so height = 1 2 4 = 1 2

i.e. the c ordinate is 1 2 or -1 2

c is above the x-axis.

So ordinate = 1 2

So y=x -2x-3=1 2

2x²-4x-7=0

x=(2±3√2)/2

So c[(2+3 2) 2,1 2] or c[(2-3 2) 2,1 2].

Hope it helps.

I wish the landlord unlimited money and everything!

The height is 5y=x2-2x-3=(x-3)(x+1), and the coordinates of a and b are (3,0)(-1,0).

then ab length = 4

The absolute value of the ordinate of point C = the height of the AB side in the triangle ABC = 10*2 4=5 substitute y=5 and y=-5.

5=x-square-2x-3

x-square-2x-8=0

x-4)(x+2)=0

x=4 or x=-2

i.e. (4,5) or (-2,5).

5=x-square-2x-3

x-square-2x+2=0 bsquare-4ac<0 No solution.

The high is 5Explain that the ordinate of point c is 5

And point c is on this quadratic curve. Two x's can be found

There are two locations for point C.

= =Isn't height the ordinate? Just substitute it into the parabolic equation and find the coordinates as (4,5) or (-2,5).

High is 5, that's right.

Because the problem has a point c above the x-axis, there are two solutions.

If the height is 5, the y value of point C is 5, then substitute it into y=x 2-2x-3, x 2-2x-3=(x-1) 2-4=5, and get x=4 or -2, so the coordinates of point c are (4,5) or (-2,5).