It is known that the parabola y=ax2+bx passes through point A (-3, -3) and point P (t, 0), t ≠ 0.

(1) If the parabola axis of symmetry passes through point A, as shown in the figure, please point out the minimum value of y at this time by observing the image and write the value of t;

(2) If t=-4, find the values of A and B, and point out the opening direction of the parabola at this time;

(3) Write directly the t value that makes the parabolic opening downward.

answer

Solution: (1)∵ Parabolic symmetry axis passes through point A,

Point ∴A is the vertex of a parabola,

The minimum value of ∴y is -3,

Point p and point o are symmetrical,

∴t=-6；

(2) Substitute (-4,0) and (-3,3) into y=ax2+bx respectively to obtain:

16a-4b=0

9a-3b=-3

Solve,

a= 1

b=4

∴ Parabolic opening direction upward;

(3) substitute A(-3, -3) and point P(t, 0) into y=ax2+bx,

9a-3b=-3①

at2+bt=0②

From ①, b=3a+ 1③,

Substituting ③ into ② gives at2+t(3a+ 1)=0.

∵t≠0,∴at+3a+ 1=0，

∴a=-

1

t+3

.

∵ Parabolic opening downward, ∴ A < 0,

∴-

1

t+3

0，

∴t>-3.

So the value of t can be-1 (the answer is not unique).

(Note: Writing t >-3 and t≠0 or any of them will get extra points)

analyse

(1) As can be seen from the figure, point A is the vertex of the parabola, and the opening is upward, so this point is the minimum value of this function;

(2) Point P is the intersection of parabola and X axis, and the other intersection is 0, so P and O are two symmetrical points about the axis of symmetry of parabola. Knowing the coordinates of the symmetrical point, it is easy to find the value of t;

(3) When a > 0, the opening of parabola is upward, and when a < 0, the opening of parabola is downward. Find the value of a and you will know its opening direction.

This topic mainly investigates the symmetry and opening direction of parabola, and is familiar with the images and properties of quadratic function.