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Ch.12 - Parametric and Polar Curves
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 12, Problem 12.4.35

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)

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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.
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 \(y = -6\).
Write the equation after substitution: \(-6 = a \times (2)^2\).
Solve for \(a\) by dividing both sides of the equation by \(4\) (since \(2^2 = 4\)), which gives \(a = \frac{-6}{4}\).
Write the final equation of the parabola by substituting the value of \(a\) back into the general form: \(y = ax^2\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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).
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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.
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Critical Points

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.
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Disk Method Using y-Axis