Find the range of function y = - 2x square + 3x + 5x ∈ R

Find the range of function y = - 2x square + 3x + 5x ∈ R

y=-2x^2+3x+5
= -2(x^2-3/2x+9/16)+49/8
=-2(x-3/4)^2+49/8
（-∞，6.125]
We can use this quadratic function formula to get the
y=-2x^2+3x+5
= -2(x^2-3/2x+9/16)+49/8
=-2(x-3/4)^2+49/8
Find the range of function y = 2x square - 3x + 5 x ∈ [- 1,1]
y=2x²-3x+5
=2(x²-3x/2+9/16)-9/8+5
=2(x-3/4)²+31/8
The axis of symmetry is x = 3 / 4 with the opening upward
therefore
When x = 3 / 4, the minimum value is 31 / 8
When x = - 1, the maximum value is 10
So the range is [31 / 8,10]
Find the range of the square + 3x + 5 of the function y = 2x in the following interval (1) x ∈ (- ∞, 1]
Y=2x^2+3x+5
Opening upward, axis of symmetry x = - 3 / 4
The axis of symmetry is in the interval (- ∞, 1]
The minimum value is the extreme value: F (- 3 / 4) = 2 * (- 3 / 4) ^ 2 + 3 * (- 3 / 4) + 5 = 9 / 8-9 / 4 + 5 = 31 / 8
Range [31 / 8, + ∞)
(x + y) (X-Y) + (X-Y) square - (X-Y) (3x-y) where x = - 2, y = 1 / 3
(x + y) (X-Y) + (X-Y) square - (X-Y) (3x-y)
=(x-y)(x+y+x-y-3x+y)
=(x-y)(-x+y)
=-(x-y)²
=-(- 2-1 / 3) & sup2;
=-(- 7 / 3) & sup2;
=-49 out of 9
(x^2-y^2)+x^2+y^2-2xy-(3x^2-4xy+y^2)
=-x^2-y^2+2xy
=-(x-y)^2
=-(-5)^2=-25
(x+y)(x-y)+(x-y)²-(x-y)(3x-y)
=（x－y）（x＋y＋x－y－3x＋y）
=－（x－y）²
When x = - 2, y = 1 / 3, the original = - (- 2-1 / 3) & sup2; = - 49 / 9
-25
What is the range of y = 3x / (- 3 | of the square of | x)?
And y = (the square of 9-x under the root) / (the square of X-9)?
[-√3/2,√3/2]
X is not equal to 3, X is a real number
The range of F (x) = the square of 3x - 5x + 2
Steps to take
y=3x^2-5x+2
=3(x^2-5/3x)+2
=3(x-5/6)^2-1/12
So y ≥ - 1 / 12
Range: ≥ 47 / 36
f(x)=3x^2-5x+2 =3(x^2-5x/3+25/36-25/36)+2=3(x-5/6)^2-1/12
When x = 5 / 6, the minimum value of F (x) = - 1 / 12, that is, f (x) ≥ - 1 / 12
The square of y = 2x + 3x + 7x belongs to the range of [2.7]
The range of this function is 21126
Please accept
Y = 2 (the square of X + 3 / 2x + 9 / 16) - 9 / 8 + 7 = 2 (x + 3 / 4) square + 47 / 8, and X belongs to [2.7]. When x = 2 has the minimum value, y = 21; when x = 7 has the maximum value, y = 126, so the range is [21.126]
What is the range y = 3x / (X & # 178; + X + 1)?
The range of y = 3x + 1 / X-1
Y = [3 (x-1) + 4] / (x-1) = 3 + 4 / (x-1), because 4 / (x-1) is not equal to 0, so y is not equal to 3,
The range is (- ∞, 3) ∪ (3, + ∞)
In general, the function y = (AX + b) / (Cx + D) (where a, B, C and D are constants and C is not equal to 0, a and B are not all 0)
The range of the function is (- ∞, a / C) ∪ (A / C, + ∞)
Y = 3x + 1 / X-1 = (3x-3 + 4) / X-1 = 3 + 4 / X-1, the range is {y ≠ 3}
Real numbers not equal to 1
Y = (x ^ 2-3x + 2) / (x ^ 2 + X + 1)