31–38. Equations of parabolas Find an equation of the following parabolas. Unless otherwise specified, assume the vertex is at the origin.
A parabola symmetric about the y-axis that passes through the point (2, -6)

Verified step by step guidance

1. Identify the general form of the parabola symmetric about the y-axis with vertex at the origin. This form is given by \(y = ax^2\), where \(a\) is a constant to be determined.
2. Substitute the coordinates of the given point \((2, -6)\) into the equation \(y = ax^2\) to find the value of \(a\). This means plugging in \(x = 2\) and \(y = -6\).
3. Write the equation after substitution: \(-6 = a \times (2)^2\).
4. Solve for \(a\) by dividing both sides of the equation by \$4\( (since \)2^2 = 4\(), which gives \)a = \frac{-6}{4}$.
5. Write the final equation of the parabola by substituting the value of \(a\) back into the general form: \(y = ax^2\).

Key Concepts

Standard Form of a Parabola

A parabola symmetric about the y-axis with its vertex at the origin can be expressed as y = ax². This form shows that y depends on the square of x, and the coefficient a determines the parabola's width and direction (upward if a > 0, downward if a < 0).

Using a Point to Find the Coefficient

To find the specific equation of a parabola, substitute the coordinates of a known point on the curve into the standard form. This allows solving for the coefficient a, tailoring the general equation to the given parabola.

Symmetry About the y-Axis

A parabola symmetric about the y-axis means its graph is mirrored on either side of the y-axis. This symmetry implies the equation involves only even powers of x, ensuring that f(x) = f(-x) for all x.