Math Quadratic Functions 10, Math Quadratic Functions

2) Proof: From (1), the equation of the parabola is y=x -2x, and the axis of symmetry is the straight line x=1

Let p be on the parabola and to the right of the axis of symmetry, let p(p,p -2p)(p>1), and let the equation for the line op be:

y=k1x, substituting the coordinates of p, has:

p²-2p=k1p

Solution: k1=p-2

Therefore, the equation for the straight line op is:

y=(p-2)x

Let x=1, get y=p-2, so the coordinates of point b are (1, p-2).

Let the coordinates of c be (1,c), and the two points b and c are symmetrical with respect to a(1,-1), then there is:

c+p-2)/2=-1

Solution: c=-p

That is, the coordinates of point c are (1,-p).

Let the equation of the straight line cp be y=k2x+b, and substitute the coordinates of c and p respectively to obtain :

p=k2+b

p²-2p=k2p+b

Solution: k2=p, b=-2p

So, the equation for the straight line cp is y=px-2p

Let y=0, from p>1, x=2, so the straight line cp must pass through the point d(2,0).

That is, the three points p, c, and d(2,0) are on the same line, and the point d(2,0) is exactly the intersection point of the parabola y=x-2x and the x-axis in addition to the origin, and o, d must be symmetrical with respect to the straight line x=1, so there must be pcb= ocb

3) When the vertices of the parabola move along the straight line y=-x, after n translations (1 n 12), the vertex coordinates are ai(1+n-1,-1-(n-1)), that is, an(n,-n), so the coordinates of the other three points of the square are dn(n,0),en(2n,0),fn(2n,-n), and the parabolic equation becomes: y+n=(x-n).

Suppose a point fn is exactly on the parabola y+i=(x-i) (where 1 i 12 and i is an integer), then there is:

n+i=(2n-i)²

n=[(4i-1)±√8i+1)]/8

When i=2,4,5,7,8,9,11,12, (8i+1) are irrational numbers, which does not match the topic;

When i=1, n=0 or 3 4 is not in line with the topic;

When i=3, get n=2 or 3 4 (rounded);

When i=6, get n=2 or 15 4 (rounded);

When i=10, n=6 or 15 4 (rounded) is not in line with the topic.

In summary, when n=2, the point f2(4,-2) is on the parabola y+3=(x-3) and y+6=(x-6) at the same time, and the square side length is 2

When n=6, the point f6(12,-6) is on the parabola y+10=(x-10), and the square side length is 6

1) From the inscription, it can be seen that the parabola and the x-axis intersect (0,0) and (2,0), and y=x -2, a=1 is obtained by substituting the original formula

2) Cross point A and point B respectively to make parallel lines on the X axis, and intersect the extension lines of OC and PC respectively at points D, E and points F and G, and connect BC, DE, and FG

BC on the axis of symmetry, de BC

Ba=AC BC is the perpendicular bisector of FCG.

Hence pcb= ocb

The coordinates of the fixed point are (3,1) before serving the potato for the difference of movement

Using the vertex formula, the analytic formula of the quadratic function is .

f(x)=a(x-3) 2 +1, the quadratic coefficient of this problem is 1, so a=1

So f(x) virtual = x 2-6x+10

The vertices of the original function can be translated back in reverse.

So the original vertex is (3,1).

With the vertex formula, let the expression Patience be y=(x-3) +1=x Changjing hail orange-6x+10