Find an equation of quadratic function the graph passing through the pointers A(2,3) B(-1,0) C(1,0) and draw the parabola

Answer 1

We can use the fact that a quadratic function can be written in the form:

y = ax^2 + bx + c

where x and y are the variables, and a, b, and c are constants.

To find these constants, we can use the given points.

Using point A(2,3), we get:

3 = a(2)^2 + b(2) + c
3 = 4a + 2b + c (equation 1)

Using point B(-1,0), we get:

0 = a(-1)^2 + b(-1) + c
0 = a - b + c (equation 2)

Using point C(1,0), we get:

0 = a(1)^2 + b(1) + c
0 = a + b + c (equation 3)

Now we have three equations with three unknowns (a, b, and c). We can solve this system of equations using substitution or elimination methods. Here we will use elimination.

Subtracting equation 2 from equation 1, we get:

3 = 4a + 2b + c
-0 = -a + b - c
-----------------
3 = 3a + 3b (equation 4)

Subtracting equation 3 from equation 2, we get:

0 = a - b + c
-0 = a + b + c
---------------
0 = 2a (equation 5)

Now we can solve for a:

equation 5 gives: 0 = 2a => a = 0

Substituting a = 0 into equation 4, we get:

3 = 3b => b = 1

Substituting a = 0 and b = 1 into equation 1, we get:

3 = 0(2)^2 + 1(2) + c => c = 1

Therefore, the equation of the quadratic function is:

y = x+1

To draw the parabola, we can use the vertex form of the equation:

y = a(x - h)^2 + k

where (h,k) is the vertex of the parabola, and a is the same constant as before (in this case, a = 0).

The vertex is at the point (-b/2a, c - b^2/4a), which we can find using the coefficients we just found:

vertex = (-1/(2*0), 1 - 1/(4*0)) = (undefined, 1)

This means that the parabola is a vertical line passing through the point (undefined, 1). We can also see this from the equation y = x + 1, which is a linear function with slope 1 and y-intercept 1.

Here is a graph of the three given points and the line y = x + 1: