It is known that a parabola is obtained by translating the parabola y = ax & # 178; upward by three unit lengths, and the parabola passes through the point (1,0), and its function analytic formula is obtained

It is known that a parabola is obtained by translating the parabola y = ax & # 178; upward by three unit lengths, and the parabola passes through the point (1,0), and its function analytic formula is obtained

A parabola is obtained by translating the parabola y = AX2 upward by 3 units
So y = ax ^ 2 + 3
And the parabola goes through the point (1,0),
therefore
0=a+3
a=-3
The function relation of parabola is y = - 3x ^ 2 + 3

Given that the image of quadratic function f (x) passes through three points a (1, - 6) B (- 1,0) C (2.5,0), the analytic expression of F (x) is obtained
Sorry, I've worked it out. a=2 b=-3 c=-5 2X^2-3X-5=0

Write out the formula of quadratic function
y=ax²+bx+c （a≠0）
Then, we substitute a, B and C into three points and calculate a, B and C respectively
①：-6=a+b+c ②：0=a-b+c ③：0=6.25a+2.5b+c
A = b = C can be solved by simultaneous methods=

If the image of a quadratic function is known to pass through three points a (0,0), B (1,3), C (3, - 6), the analytic expression of the quadratic function is obtained

Let the function be y = ax ^ 2 + BX + C
I've been to ABC, so I have
0=c
3=a+b+c
-6=9a+3b+c
The solution is a = - 5 / 2, B = 11 / 2, C = 0
So the analytic formula is y = - 5 / 2x ^ 2 + 11 / 2x

If the image of a quadratic function is known to pass through points a (1, - 1), B (5, - 1), C (3,3), the analytic expression of the quadratic function is obtained

y=-(x-3)^2+|3

How do quadratic functions know whether a, B, C are greater than 0 or equal to 0 or less than 0

Just look at the picture
1. Look at the opening first, mouth up - A is greater than 0
Opening down - A is less than 0
2. Look at the position of the axis of symmetry
The axis of symmetry is on the left side of the x-axis
The axis of symmetry is on the right side of the x-axis
We already know whether a is greater than or less than 0
Use this to know the positive and negative of B
3. The intersection of image on Y axis
Intersection on positive semiaxis -- C greater than 0
Intersection on negative semiaxis -- C less than 0
Intersection at origin -- C equals 0

Why can we judge a > 0 when quadratic function is greater than or equal to zero

Because if the quadratic function is greater than or equal to 0, the opening must be upward, then a must be greater than 0;
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In quadratic function, for example, if a + B + C is proved to be greater than 0, then we can say x = 1. How do we know if y is greater than 0 when x = 1 is substituted into quadratic function

In fact, this kind of problem, often through the image to observe whether the corresponding y value is positive when x = 1

How to judge whether AC, a + B + C, 4a-2b + C, 2A + B, 2a-b are greater than 0?

Find a specific value that meets the conditions and substitute it to judge

It is known that the maximum value of quadratic function f (x) is 8, and its image passes through two points a (- 2,0) and B (1,6), then the analytic expression of the function is

Let f (x) = a (X-B) ^ 2 + 8 (a)

Given the coordinates of ABC three points on the image of quadratic function, find the analytic expression a (1 / 2, - 1 / 2) B (0,2) C (1, - 5) of this function

Let y = ax & # 178; + BX + C, bring the points a, B, C into the following result:
-1/2=ax（1/2）²+bx1/2+c
2=ax0²+bx0+c
-5=ax1²+bx1+c
To solve this system of equations, a = - 4, B = - 3, C = 2