Quadratic function questions for junior high school, quadratic function questions for junior high sc

1) Proof: m 4 (1) (7 2m) m 8m 28 (m 4) 12 0

The parabola has 2 points of intersection with the x-axis.

2) The parabola has 2 points of intersection with the x-axis: a(x1,0)b(x2,0)x1 x2 m x1x2 2m 7

ab＝|x2－x1| ∴ab²＝|x2－x1|²＝x1＋x2)²－4x1x2

m 8m 28 16 m 8m 12 0 m 2 or 6 The intersection of the parabola and the y-axis is on the positive semi-axis 7 2m 0 m 7 2m 2 The parabola is: y x 2x 3

1.△＝b²-4ac=m²+4(7-2m)=m²-2m+14=(m-2)²+13＞0

Regardless of the value of m 0, the parabola and the x-axis always have two intersections.

x2-x1=4 according to the title

x2+x1=m

x2*x1=-(7-2m)

Lianli solution. m-2)(m-6)=0

m=2 or.

m=6 The intersection of the parabola and the y-axis is rounded off at the positive semi-axis 7 2m 0 m 7 2m=6.

So parabola.

y= -x²+x+3

Let the time be t and the distance be y

Then y = 12t +26-5t =169t -260t+676 When time t = 260 338 = 10 13, y = 169 (10 13) -260 10 13 + 676 = 576

The closest distance is y=24

Let x be the closest when you are an hour.

12x) squared + (26-5x) squared.

144x2+26 squared - 260x+25x2=169x2-260x+26 squared.

13x-10) square + 26 square - 100

Therefore, if it is equal to 0 in the broad sign, it is the smallest, which is equal to 26 square meters - 100

If all limbs pass through different points on the x-axis, the axis of symmetry should be the same.

a=(a-3)/2 a=1

x +2ax-2b+1=0 -x +(a-3)x+b -1=0 has a solution that is not the same in the phase.

a=1, b=2, or b=0

1>0 2> Lipeidong 0

b=0 rounded, b=2

In summary, a=1 b=2

From the problem, we know that the equation x 2 + 2ax-2b + 1 = 0 and the equation x 2 + (3-a) x + 1-b 2 0 have the same solution.

According to the two root relationships, it can be destroyed:

2a＝3－a

b^2－1＝2b－1

So, a 1, b 0 or b 2

Because, when b 0, the equation x 2 + 2ax - 2b + 1 = 0 has only one intersection point with the x-axis, so it is discarded.

So, a 1, b 2

The axis of symmetry is the straight line x=-1, i.e. -b (2a) = -1, so b = 2a a 0 b>a

Let the other intersection of the parabola and the x-axis be x2, then x1+x2=-2, and since 0 x1 1, -3 x2 -2

The parabolic opening is upward, and it can be seen from the image that the function value of x=1 and -3 should be greater than 0.

That is: a+b+c>0......①

9a-3b+c>0……②

3+ gets: 12a+4c>0 i.e.: 3a+c>0

The first and third conclusions are correct, the second is wrong.

The first one is set to y=a(x-1) 2-3 and then substituted the point (0,1).

The second wisdom is set y=a(x+3)(x-5) and then the point before the uproar (reeds 0,-3).

The third let y=a(x-3) 2-2 have a distance of 4, which means that two units on the two sides of the axis of symmetry x=3 can be (1,0) or (5,0).