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Magazine Chapter 3: Problem 32
Exer. 29-34: Find the standard equation of a parabola that has a vertical axis and satisfies the given conditions. Vertex \((4,-7), x\)-intercept \(-4\)
Short Answer
Expert verified
The equation is \(y = \frac{7}{64}(x - 4)^2 - 7\).
Step by step solution
01
Understand the Problem
We need to find the standard equation of a parabola with a vertical axis that passes through a given vertex and has a specified x-intercept. The vertex of the parabola is given as \((4, -7)\), and it passes through the x-intercept \(-4\).
02
Write the Equation of a Parabola with a Vertical Axis Using Vertex Form
The equation for a parabola with a vertical axis in vertex form is \(y = a(x - h)^2 + k\), where \((h, k)\) is the vertex of the parabola. For this problem, the vertex \((h, k)\) is \((4, -7)\). Thus, the equation becomes \(y = a(x - 4)^2 - 7\).
03
Substitute the x-intercept into the Equation to Find 'a'
An x-intercept means the point is on the parabola where \(y = 0\). For the x-intercept at \(-4\), substituting into the equation gives: \[0 = a(-4 - 4)^2 - 7\]Solve for 'a' to find: \[0 = a(64) - 7 \a \times 64 = 7 \a = \frac{7}{64} \].
04
Write the Final Standard Equation
Now that we know the value of \(a = \frac{7}{64}\), substitute \(a\) back into the vertex form equation: \[y = \frac{7}{64}(x - 4)^2 - 7\].
05
Verify the Solution
To ensure the parabola passes through the x-intercept \(-4\), substitute \(x = -4\) into the equation and confirm that \(y = 0\): \[0 = \frac{7}{64}((-4) - 4)^2 - 7\].The equation holds true, verifying our solution is correct.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vertex Form
The vertex form of a parabola provides a clear picture of its shape and orientation around its vertex point. This specific form is written as \( y = a(x - h)^2 + k \), where:
This equation shows that the parabola is centered horizontally around \(x = 4\) and vertically around \(y = -7\).
- \((h, k)\) represents the vertex of the parabola.
- \(a\) determines how wide or narrow the parabola is; if \(a\) is positive, the parabola opens upwards, and if \(a\) is negative, it opens downwards.
This equation shows that the parabola is centered horizontally around \(x = 4\) and vertically around \(y = -7\).
Vertical Axis
A parabola with a vertical axis is one where the line of symmetry runs vertically. This means the parabola will open either upwards or downwards. When we talk about a parabola having a vertical axis, we're primarily concerned with its orientation:
This indicates a standard upward-facing parabola with its turning point at the vertex \((4, -7)\).
- With the vertex form \( y = a(x - h)^2 + k \), the parabolic curve will open.
- The direction of opening is dictated by the sign of \(a\); here, a positive \(a\) means the curve opens upwards, and a negative \(a\) implies it opens downwards.
This indicates a standard upward-facing parabola with its turning point at the vertex \((4, -7)\).
x-intercept
The x-intercept of a parabola is where it crosses the x-axis, meaning the y-value at this point is zero. Knowing an x-intercept is critical for reaching the full equation as it provides additional information to solve for \(a\) in the vertex form.
- For example, if given an x-intercept \(-4\), you substitute \(-4\) for \(x\) and \(0\) for \(y\) in the vertex form.
- This yields the equation \(0 = a(-4 - 4)^2 - 7\) which simplifies to solve for \(a\).
- Solving gives \(a = \frac{7}{64}\).
Standard Equation
The standard equation of a parabola generally takes the form \( Ax^2 + Bx + Cy = 0 \), but a vertically oriented equation can also derive from the vertex form. Once \(a\) is identified, substitute it back to form the complete parabola equation:
This process confirms our derived equation is correct, ensuring our parabola behaves as expected.
- The vertex form \( y = \frac{7}{64}(x - 4)^2 - 7 \) becomes our operational base.
- This equation describes the full parabola, indicating it reaches its lowest point at the vertex (4, -7) and passes through the x-intercept at (-4, 0).
This process confirms our derived equation is correct, ensuring our parabola behaves as expected.
