If the minimum value of the quadratic function y ax 4x a 1 is 2, then the value of a is

Hello. y=ax²+4x+a-1

a（x²+4/ax+4/a²）-a*4/a²+a-1a（x+2/a）²-4/a+a-1

When x = 2 a, y has a minimum value of 2 and a 0

4/a+a-1=2

4/a+a-3=0

4+a²-3a=0

a-4）（a+1）=0

a=4, or a=-1 (rounded).

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The main thing is to forget the judgment of a positive and negative, so the conclusion is always reached, a 0, -3a-1 is not the minimum value, so it should be considered from another angle, because the minimum value of y is 2, so a 0, ax +4x + a-1 2, so ax +4x 3 a, , y=ax +4x The minimum value is 4 a, so 4 a 3 a, solve a 4, this is the final answer, I hope it will help you, if you can please choose a satisfactory answer, I urgently need your help, thank you.

y=ax²+4x+a-1

a[x²+(4/a)x+4/a²]-4/a+a-1a(x+2/a)²-4/a+a-1

Because y has a minimum value, a>0

When x=-2 a, y min=-4 a+a-1, so -4 a+a-1=2

a²-3a-4=0

a=4 or -1

and a>0, so a=-1 rounding.

Therefore a=4

ax²+4x+a-1>=2

And then you should be able to do it, right?

Solution: Quadratic function y ax 2 4x (a-1) a(x 2 a) 2 (a-4 a-1).

Quadratic functions have a minimum value, and a minimum value of 2

a＞0，a-4／a-1＝2

A-4 A-3 0, A 2-3A-4 0, (A 1) (A-4) 0, then A 4, A -1 are rounded. a＝4

y=a(x^2+4x/a)+a-1=a(x+2/a)^2+a-1-4/a

Minimum = 2 = a-1-4 a

i.e. a-3-4 a=0

a^2-3a-4=0

a-4)(a+1)=0

If there is a minimum value, then a>0, so there is a-4=0

Get: a=4

y = x square - 4y + a

5y=x flat cherry chop square + a

y = square + a

The minimum value of the friend is equal to (4 squared) (

a=50

Because it is a quadratic function, a cannot be zero, and because there is a minimum value, the parabolic opening is upward, a>0, and the axis of symmetry of the parabola is the y-axis, and the minimum value should be y=axa-4=5 when x=0

axa=9，a=3

(1) If a=0, then y=4x-1, there is no minimum value of the function.

2) If a≠0, the function is parabolic, and if the function y has a minimum value, then:

y=a(x+2 a) 2+(a 2-4-a) aa<0, and the minimum value (a 2-a-4) a=2 is solved to a=-1, a=4 (disagreement, rounded).

a=-1

The formula for the minimum value of the quadratic function is x=(4ac-b) 4a a-1-4 a=2

a-3-4/a=0

Multiply a on both sides of the equation.

a²-3a-4=0

a-4）（a=1）=0

a = 1 or 4

Isn't there a vertex formula: vertex coordinates ( -b 2a, (4ac-b 2) 4a ) substituting b=4, c=a-1 into vertices (4ac-b 2) 4a gives a=4 or a= -1, you are talking about the minimum value of 2, then the function opening is upward, i.e. a>0, so a=4 .

Because this function is quadratic and has a minimum value, the function opening is upward, i.e., a is greater than 0

The minimum value of the function is at the vertex, when x=-4 2a, y=2Substituting the value of a is 4 or -1

And because a is greater than 0, a = 4

The function has a maximum value indicating that the parabolic opening is downward, and we get a<0

In addition, the function has an extreme value when x=-b 2a, here is the maximum value that brings x=4 2a=2 a into the function y=ax 2-4x+a-1, and the y value at this time is 2.

a 2 - 3a -4 = 0 solution.

a=4 (rounded) or a=-1

Get a=-1

From the title, there is a minimum value, then a>0y=ax 2+4x+a-1=a(x+2 a) 2-4 a+a-1-4 a+a-1=2a=4 or -1 (does not meet the topic, round).

y=ax +4x+a-1=a(x+2 a) +a-1-4 a, the minimum value of a is 2, i.e., a 0, and a-1-4 a=2 simplify: a -3a-4=0 The solution gives a=4 or a=-1 (rounded), so the value of a is 4

Since it is the minimum value, then a>0 is substituted into the verification one by one. Choose A