Problem 29 - Precalculus with Limits

Question

In Exercises 19-32, find the standard form of the equation of the parabola with the given characteristic(s) and vertex at the origin. Vertical axis and passes through the point \((4, 6)\)

Step-by-Step Solution

Answer

The standard form equation of the parabola is \(y = 0.375x^2\).

Step 1: Note Down the Given Information

Vertex of the parabola is at the origin (0, 0) and the parabola passes through the point (4, 6). The standard form of the equation for a parabola with a vertical axis is \(y = ax^2\), where 'a' is a constant.

Step 2: Substitute the Given Point

Now, since the parabola passes through the point (4,6), substitute this point into the equation to solve for 'a'. It gives \(6 = a*4^2\) or \(6 = 16a\).

Step 3: Solve for 'a'

From the equation in step 2, solve for 'a'. This calculation yields \(a = 6/16\) or \(a = 0.375\).

Step 4: Write the Final Equation

Substitute the value of 'a' obtained in step 3 into the standard form equation to write the final equation of the parabola. It gives \(y = 0.375x^2\).

Key Concepts

Vertex of a Parabola

The vertex of a parabola is essentially the 'tip' of the curve, the point where it reaches its minimum or maximum value. In the context of our exercise where we have a vertical axis of symmetry and the vertex at the origin (0,0), this is exceptionally straightforward. The parabola's standard form equation being based around its vertex allows us to easily plot its course on a graph.

In this case, since the vertex is at the origin, the equation of the parabola remains centered on the coordinate plane, hence we start with the form \(y=ax^2\) where 'a' affects the width of the parabola; a higher value makes it narrower, while a lower value makes it wider.

Axis of Symmetry

The axis of symmetry in a parabolic equation is a straight line that divides the parabola into two mirror-image halves. For any parabola in standard form \(y=ax^2+bx+c\), the axis of symmetry can be derived from the quadratic coefficients and is represented by the equation \(x=-\frac{b}{2a}\).

However, when the vertex is at the origin and the equation simplifies to \(y=ax^2\), as in our exercise, the 'b' and 'c' terms are zero, affirming that the axis of symmetry is the y-axis itself, or \(x=0\). This simplification provides symmetry across the y-axis, creating a visual balance that reinforces the property of the parabolic shape.

Solving Quadratic Equations

Solving quadratic equations is a fundamental skill in algebra. The general form of a quadratic equation is \(ax^2+bx+c=0\), and its solutions can be found using various methods including factoring, completing the square, or applying the quadratic formula. But in the context of a parabola's standard form when the vertex is at the origin, the equation simplifies and the process becomes easier.

For our specific example, we are solving for 'a' when the parabola passes through a known point, like (4, 6). Here, substitution followed by basic algebraic manipulation provides us with the value of 'a'. Once 'a' is determined, the parabola's equation in standard form becomes immediately usable for graphing and further analysis - pivotal for visualizing and understanding the behavior of quadratic relationships.