Question

Find an equation for the conic section with the given properties. The parabola that passes through the point (2, 2), with vertex V(-1, 3), and a horizontal axis of symmetry.

a) y = -2x^2 + 5x - 1
b) y = x^2 + 3x - 2
c) y = -x^2 + 4x - 3
d) y = 2x^2 - 3x + 1

Answer 1

To find the equation of the given parabola, we need to substitute the coordinates of the vertex and the point it passes through into the general equation of a parabola. By solving the resulting system of equations, we can find the values of the coefficients and write the equation of the parabola.

Step-by-step explanation:

The equation of a parabola with a horizontal axis of symmetry can be written as y = ax^2 + bx + c. To find the equation that represents the given parabola, we need to substitute the coordinates of the vertex, (-1, 3), into the equation. This will give us the values of a, b, and c. Let's plug in the vertex coordinates:

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

Simplifying the equation, we get:

3 = a - b + c

We also know that the parabola passes through the point (2, 2). Let's substitute these coordinates into the equation as well:

2 = a(2)^2 + b(2) + c

Simplifying the equation, we get:

2 = 4a + 2b + c

Now, we have a system of two equations with three variables, a, b, and c. We can solve this system to find the values of a, b, and c. Once we have the values of a, b, and c, we can write the equation of the parabola.