A parabola opening up or down has vertex (0, 0) and passes through (4, 1). Write its equation in vertex form. Simplify any fractions.
Understand the Problem
The question is asking to write the equation of a parabola in vertex form given its vertex at (0, 0) and a point (4, 1) through which it passes. This involves using the vertex form of a parabola and substituting the known values into it.
Answer
The equation of the parabola is given by $$ y = \frac{1}{16}x^2 $$.
Answer for screen readers
The equation of the parabola in vertex form is: $$ y = \frac{1}{16}x^2 $$
Steps to Solve
- Identify the vertex form of a parabola
The vertex form of a parabola is given by the equation: $$ y = a(x - h)^2 + k $$ where $(h, k)$ is the vertex of the parabola.
- Substitute the vertex into the equation
Given the vertex at $(0, 0)$, we substitute $h = 0$ and $k = 0$ into the equation: $$ y = a(x - 0)^2 + 0 $$ which simplifies to: $$ y = ax^2 $$
- Use the point to find the value of $a$
We know that the parabola passes through the point $(4, 1)$. We will use this point to find $a$. Substitute $x = 4$ and $y = 1$ into the equation: $$ 1 = a(4)^2 $$ which simplifies to: $$ 1 = 16a $$
- Solve for $a$
Rearranging the equation gives us: $$ a = \frac{1}{16} $$
- Write the final equation in vertex form
Now that we have found $a$, we substitute it back into the equation: $$ y = \frac{1}{16}x^2 $$
The equation of the parabola in vertex form is: $$ y = \frac{1}{16}x^2 $$
More Information
The vertex form of a parabola allows us to easily identify its vertex, as well as how "wide" or "narrow" it opens based on the value of $a$. In this case, a small value of $a$ indicates that the parabola is relatively wide.
Tips
- Forgetting to substitute the vertex values correctly into the vertex form equation.
- Not using the correct point coordinates for solving the value of $a$.
- Miscalculating the value of $a$ when rearranging the equation.
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