Write down the coefficient of quadratic term, coefficient of first term and constant term of quadratic function Coefficient of quadratic term and constant term of coefficient of first term in analytic expression of function y=x^2+2x-1 y=x^2 y=-3x^2+2 y=1/3(x-5)^2-4

Write down the coefficient of quadratic term, coefficient of first term and constant term of quadratic function Coefficient of quadratic term and constant term of coefficient of first term in analytic expression of function y=x^2+2x-1 y=x^2 y=-3x^2+2 y=1/3(x-5)^2-4

Coefficient of quadratic term and constant term of coefficient of first term in analytic expression of function
y=x^2+2x-1 1 2 -1
y=x^2 1 0 0
y=-3x^2+2 3 0 2
y=1/3(x-5)^2-4 1/3 -10/3 13/3

Which of the following functions (x.t is an independent variable) are quadratic? Point out the coefficients of quadratic term, coefficient of first term, and constant term
1. Y = - 1 / 2 + 3x square,
2. Y = (x-3) (4-2x) + 2x square
3. S = root 5T square + T + 1,
4. Y = x square-3 radical X-7

1. Y = - 1 / 2 + 3x quadratic function, coefficient of first-order term. 3 and constant term. - 1 / 22. Y = (x-3) (4-2x) + 2x square = 10x-12, first-order function 3. S = root sign 5T square + T + 1 two kinds of explanation 1. S = root sign (5T square + T + 1) second-order radical is not a quadratic function, 2. S = root sign 5 * (T square + T + 1) is a quadratic function, first-order term

What is the general formula of the quadratic function y = (x-1) x (2-x), the coefficient of the quadratic term, the coefficient of the primary term and the constant respectively

General formula: y = ax & sup2; + BX + C, the coefficient of quadratic term is a, the coefficient of primary term is B, and the constant is C
In your question: y = (x-1) x (2-x) = - X & sup2; + 3x-2
So the coefficient of the quadratic term is - 1, the coefficient of the primary term is 3, and the constant is - 2

If y = (X-2) (3-x) is a quadratic function, write out its quadratic coefficient and its linear coefficient constant

Yes, the coefficient of quadratic term is - 1, the coefficient of primary term is 5, and the constant term is - 6

Write out a quadratic function relation, make its image pass through point (1,0), and the coefficient of quadratic term is 1

y=x^2-1

The coefficient of the quadratic term of a quadratic function is 1 and the coefficient of the first term is 0. The coordinates of the intersection of the image of the quadratic function and the Y axis are (0,1). The analytic expression of the quadratic function is

Let y = x ^ 2 + B
Substituting (0,1)
b=1
y=x^2+1

When drawing the image of quadratic function with point tracing method, some data are shown in the following table
x ··· -2 -1 0 1 2 3 ···
y ··· 9 4 1 0 1 4 ···
Find the relation of quadratic function corresponding to the image

Let the quadratic function be y = ax & # 178; + BX + C
Substituting (0,1), (1,0), (2,1) respectively, we get the following results:
{1=c
{0=a+b+c
{1=4a+2b+c
The solution is: a = 1, B = - 2, C = 1
∴y=x²-2x+1

The image of quadratic function passes through the origin, and f (- 2) = 0, f (1) = 3. Find the analytic expression of quadratic function, and find the vertex coordinates

Let the analytic expression of the quadratic function be y = ax & sup2; + BX + C. according to the meaning of the problem, the points (0,0), (- 2,0) and (1,3) are substituted into
{c=0
4a－2b＋c=0
a＋b＋c=3
The solution is {A1}=
b=2
c=0
Therefore, the analytic expression of quadratic function is y = x & sup2; + 2x
∵y=x²＋2x=(x＋1)²－1
The vertex coordinates are (- 1, - 1)

Help to do a quadratic function diagram to high score
The image of quadratic function y = ax & sup2; + BX + C and the image of quadratic function y = 2x & sup2; have the same shape, size, opening direction and vertex coordinates (- 1,2). Find the values of a, B, C

Because of the same shape, size and opening direction, a = 2, and because of the vertex (- 1,2), so - B / 2A = - 1, we can get b = 4, and substitute (- 1,2) into y = 2x2 + 4x + C, we can get C = 4

Find the analytic expression of quadratic function satisfying the following conditions: (1) the image passes through a (- 1,3), B (1,3), C (2,6); and
(2) The image passes through a (- 1,0), B (3,0), and the function has a minimum value of - 8;
(3) The vertex coordinates of the image are (- 1,9), and the ordinate of the intersection point with the Y axis is 5

Let the quadratic function be y = ax ^ 2 + BX + C1, passing through points a, B, C, so: 3 = A-B + C (1)3=a+b+c…………………… (2)6=4a+2b+c………………… (3) (2) - (1), there are: 2B = 0, the solution is: B = 0 substituting (2), (3), there are: 3 = a + C (4)6=4a+c………… (5...