A quadratic function problem for the third year of junior high school, and a quadratic function ques

Substituting the point q(0,-3) into the parabola y=x 2+bx+c, we get c=-3, and let a(x1,0) and b(x2,0).

Then the area of the triangle PAB = x1-x2 * 12-b 2) 4 2, so x1-x2 * 12-b 2) 4 = 16 root number[(x1+x2) 2-4x1x2]*(12+b 2)=64, i.e. [root number(b 2+12)]*b 2+12) = 64, so root number(b 2+12) 3=64

So b 2 + 12 = 16, the solution is b = -2 or b = 2 (rounded) and the analytic formula is y=x 2-2x-3

Bringing the Q coordinates into the parabola equation yields -3=0+0+C to obtain C=-3, and the intersection points a,b with the x-axis can be known as x 2+bx-3=0, and the coordinates of the q point know that the abscissa of a,b is positive and negative, x(a)-x(b)=8*2 p

p represents the absolute value of its ordinate.

There is a formula for the maximum value of the ordinate in the book.

Coupled with the @ formula below.

x(a)+x(b)=-b,x(a)-x(b)] =b 2+12@ Finally, the analytic formula for b is obtained.

That's all I can say, I've almost forgotten about it, I'm sorry.

Because the quadratic function is determined by two points where it intersects the x-axis.

Therefore, the intersection function determination method is set to y=a(x- x1) (x- x2) to substitute the x in (-1,0)(4,0) respectively.

y= a[x-(-1)] x-4) = a(x+1)(x-4) because his shape is the same as y=x2, so their openings are equal in size, so the value of a is also equal, and finally the analytical formula of the quadratic function is y=-(x+1)(x-4).

If the image of a quadratic function intersects the x-axis at the points (-1,0),(4,0), we can see that the axis of symmetry of this parabola is x=

Then substitute x= into y=-x 2 to get the coordinates of another point (here there should be positive and negative solutions, and the final parabola shape is the same, but the opening direction, one up and one down).

Finally, the parabola can be found from the coordinates of the three points.

Untie; 1, y=x-2x-3, intersects a(-1,0)b(, intersects y-axis c(0,-3), vertex coordinates m(1,-4), s quadrilateral ocmb=s obm+s ocm, abrasion = 1 2ob4+1 2oc 1 because ob=oc=3. So the s quadrilateral ocmb=6.

2. Let d(m,y),(x 0,y 0) then s quadruple this reputation shape ocdb=1 2ob(-y)+1 2ocx=3 2(m-y), because d is on the parabolic blind ridge, y=m -2m-3, so the s quadrilateral ocdb=-3 2(m -3m-3)=-3 2(m-3 2) +63 8So there is a fourth quadrant where the point d maximizes the area of the quadrilateral ocdb, and the abscissa of d is m=3 2....

3, (1) Let d(m, n), because de x-axis, so de=-n=-m +2m+3,, the same way pe=3-m, pd=d=de-pe=-m +2m+3-3+m=-m +3m(2) Obviously, pd oc, if the quadrilateral dpoc is a parallelogram, only pd=oc=3, that is, m -3m+3 = 0, since the discriminant value of the root of the equation is 9-12=-3 0, d does not exist.