Problem 44 - Precalculus with Calculus Previews

Question

Find an equation of parabola that satisfies the given conditions. Vertex \((0,0),\) through \(\left(1, \frac{1}{4}\right)\), axis along the \(x-\) axis

Step-by-Step Solution

Answer

The equation of the parabola is \(y = \frac{1}{4}x^2\).

Step 1: Understand the Parabola Form

Given that the vertex is at \((0,0)\) and the axis of the parabola is along the \(x\)-axis, the equation of the parabola can be written in the form \(y = ax^2\). This is because the parabola opens either to the left or right, and its vertex form simplifies to just \(ax^2\) given that \(h=0\) and \(k=0\) in the vertex formula \(y = a(x-h)^2 + k\).

Step 2: Substitute the Point into the Equation

The parabola passes through the point \(\left(1, \frac{1}{4}\right)\). Substitute \(x = 1\) and \(y = \frac{1}{4}\) into the equation \(y = ax^2\) to solve for \(a\):\[ \frac{1}{4} = a(1)^2 \] This simplifies to:\[ \frac{1}{4} = a \]

Step 3: Write the Parabola Equation

Now that we have determined \(a = \frac{1}{4}\), substitute \(a\) back into the equation \(y = ax^2\) to find the specific equation of the parabola:\[ y = \frac{1}{4}x^2 \]

Key Concepts

Vertex Form

The vertex form of a parabola is a neat way to express the quadratic equation, where you can immediately see the vertex. This form is given by \[ y = a(x-h)^2 + k \] where

• \( (h, k) \) represents the vertex of the parabola, and
• \( a \) is the coefficient that affects the opening direction and width of the parabola.

When a parabola's vertex is at the origin, \( (0, 0) \), the vertex form simplifies to \( y = ax^2 \) or \( x = ay^2 \) based on the axis of symmetry. In our exercise, the vertex is at the origin, and the axis is along the \( x \)-axis, hence the equation simplifies to \( y = ax^2 \). This makes it easier to solve for \( a \) when given a point the parabola passes through.

Axis of Parabola

The axis of a parabola defines its line of symmetry and determines the direction in which it opens. For any parabolic equation, the axis can either be vertical or horizontal:

• If the parabola opens upwards or downwards, as with \( y = ax^2 \), the axis of symmetry is a vertical line, usually the \( y \)-axis or \( x = h \).
• If it opens to the left or right, like in our example with \( x = ay^2 \), the axis of symmetry is a horizontal line, usually the \( x \)-axis or \( y = k \).

In this exercise, with the parabola's axis lying along the \( x \)-axis and a vertex at \( (0,0) \), it horizontally opens. Knowing the axis helps us choose the appropriate form of the equation (like \( y = ax^2 \)) to solve problems easily.

Solving for Coefficients

To find the specific equation of a parabola, you often need to solve for coefficients such as \( a \), which dictate the shape and position of the curve. In our example, the parabola passes through the point \( (1, \frac{1}{4}) \). We use this point to substitute into the simplified vertex form \( y = ax^2 \) because the vertex is at \( (0,0) \).Here's how:1. Substitute \( x = 1 \) and \( y = \frac{1}{4} \) into \( y = ax^2 \).2. This gives us \( \frac{1}{4} = a(1)^2 \).3. Simplify to find \( a = \frac{1}{4} \).
Finally, you'd plug \( a = \frac{1}{4} \) back into the equation, giving you \( y = \frac{1}{4}x^2 \). This step-by-step substitution process is crucial for accurately determining the parabola's equation.