1. Write a quadratic function in intercept form whose graph has the given x-intercepts and passes through the given point.

x-intercepts: 2,3
point: (4,2)

2. Write a quadratic function in standard form whose graph passes through the given points.

(1,-2),(-2,1),(3,6)

first one is easy,

you know that x-2 and x-3 must be factors of the quadratic.
let the equation be
y = a(x-2)(x-3)
sub in (4,2) to solve for a, and you are done

the second takes a bit more work.
let the equation be
y = ax^2 + bx + c

now take each point and sub into that, giving you 3 equation in 3 unknowns.
first one:
-2 = a + b + c

take it from there.

1. To write a quadratic function in intercept form (also known as factored form), we start with the x-intercepts and the vertex.

The x-intercepts are given as 2 and 3. So, the factored form of the quadratic function is (x-2)(x-3) = 0.

Next, we need to find the y-intercept or the third point. Since the quadratic function graph passes through (4, 2), we can substitute these values in the factored form to find the constant term.

Substituting x=4 and y=2 into (x-2)(x-3) = 0, we get (4-2)(4-3) = 2(1) = 2.

Therefore, the quadratic function in intercept form is: f(x) = (x-2)(x-3)

2. To write a quadratic function in standard form, we can use the given points. The standard form of a quadratic function is f(x) = ax^2 + bx + c.

We can substitute each given point into this equation to create a system of equations.
Using the first point (1, -2):
-2 = a(1)^2 + b(1) + c --(1)

Using the second point (-2, 1):
1 = a(-2)^2 + b(-2) + c --(2)

Using the third point (3, 6):
6 = a(3)^2 + b(3) + c --(3)

Expanding and simplifying the equations, we get:
1a + 1b + c = -2 --(4)
4a - 2b + c = 1 --(5)
9a + 3b + c = 6 --(6)

Now, we can solve this system of equations to find the values of a, b, and c.

To solve equations (4) and (5), we can multiply (4) by -2 and (5) by 1, and then add both equations together:
-2a - 2b - 2c = 4
4a - 2b + c = 1
---------------
2a - c = 5 --(7)

Next, we can subtract equation (7) from equation (6):
9a + 3b + c = 6
2a - c = 5
-----------------
7a + 3b = 1 --(8)

Now, we can solve equations (7) and (8) for a and b:
2a - c = 5 --(7)
7a + 3b = 1 --(8)

Solving equation (7) for c, we get:
c = 2a - 5

Substituting this value of c into equation (8), we get:
7a + 3b = 1
7a + 3b = 2a - 5
5a + 3b = -5

Solving this new equation for b, we get:
3b = -5 - 5a
b = (-5 - 5a) / 3

Now, we have expressions for c and b in terms of a. We substitute these expressions in equation (4) to eliminate b and c:
1a + 1b + c = -2 --(4)
a + ((-5 - 5a) / 3) + (2a - 5) = -2

Simplifying this equation, we get:
a - 5 - 5a + 6a - 15 = -6
2a - 20 = -6
2a = 14
a = 7

Substituting the value of a into our expression for b:
b = (-5 - 5a) / 3
b = (-5 - 5(7)) / 3
b = (-5 - 35) / 3
b = -40 / 3

We can substitute the values of a, b, and c in the standard form equation to get the final quadratic function:
f(x) = ax^2 + bx + c
f(x) = 7x^2 - (40/3)x + (2(7) - 5)
f(x) = 7x^2 - (40/3)x + 9

To write a quadratic function in intercept form that passes through the given x-intercepts and the point, we can start by using the factored form of a quadratic function:

f(x) = a(x - r)(x - s)

where r and s are the x-intercepts, and a is a constant factor. In this case, the x-intercepts are 2 and 3, so the function will be in the form:

f(x) = a(x - 2)(x - 3)

Next, we need to find the value of a. To do this, we can substitute the coordinates of the given point into our function. The point given is (4,2), so we have:

2 = a(4 - 2)(4 - 3)

Simplifying further:

2 = 2a(1)

Now, we can solve for a:

2 = 2a
a = 1

Substituting this value of a back into our function, we get the quadratic function in intercept form:

f(x) = (x - 2)(x - 3)

To write a quadratic function in standard form that passes through the given points, we can use the general form of a quadratic equation:

f(x) = ax^2 + bx + c

We know that the quadratic function passes through points (1, -2), (-2, 1), and (3, 6). We can substitute these points into the equation to create a system of equations and solve for the unknowns a, b, and c.

For the point (1, -2):

-2 = a(1)^2 + b(1) + c --> -2 = a + b + c

For the point (-2, 1):

1 = a(-2)^2 + b(-2) + c --> 1 = 4a - 2b + c

For the point (3, 6):

6 = a(3)^2 + b(3) + c --> 6 = 9a + 3b + c

Now we have a system of three equations. To solve, we can use techniques such as substitution or elimination to find the values of a, b, and c that satisfy all three equations.