## Known quadratic function y = - 2x ² How to translate this function image so that it can pass through two points (0,0) and (1,6) Known quadratic function y = - 2x ², How to translate this function image to make it pass through two points (0,0) and (1,6), so that all experts can write a method that can be understood,

After translation, the function becomes: y = - 2 (x + a) ²+ b

It is brought in through points (0,0), (1,6):

-2a ²+ b=0

-2(1+a) ²+ b=6

The solution is a = - 2, B = 8

After translation, the function becomes: y = - 2 (X-2) ²+ eight

That is, first shift the function to the right by 2 units, and then up by 8 units

### Set the quadratic function y = X ²- The graph of 2x + 1 shifts up two units and then left three units to obtain the quadratic function y = X ²+ BX + C image, find the values of B and C

y=x^2-2x+1=(x-1)^2

Translate two units up

have to

y-2=(x-1)^2

Shift three units to the left

have to

y-2=(x-1+3)^2=(x+2)^2=x^2+4x+4

y=x^2+4x+6

b=4 c=6

### Quadratic function y = - 2x ²- If there are two intersections between X + M-1 and X axis, the value range of M is

△ = bsquare -4ac = 1 + 8 (m-1) > 0

8m>7

m>7/8

Then the value range of M is (7 / 8, + ∞)

### Known quadratic function y = x ²+ 2X + C if - 2 ＜ x ＜ 1, the quadratic function has and has only one intersection with the X axis to find the value range of C

If the Y and X axes have 2 intersections, then:

y(-2)=4-4+c=c

y(1)=1+2+c=3+c

Y (- 2) y (1)

### Quadratic function y = - x ²+ 2X + 3, if y ≥ 3, the value range of X

If y ≥ 3

Then - x ²+ 2x+3≥3

I.e. - x ²+ 2x≥0

x ²- 2x≤0

x(x-2)≤0

Therefore, the value range of X is 0 ≤ x ≤ 2

### In quadratic function y = - x ²＋ In the 2x + 1 image, if y increases with the increase of X, what is the value range of X? Explain the content.

The opening of the function is downward

The axis of symmetry is x = 1

On the left side of the symmetry axis, y increases with the increase of X

So the value range of X is X