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Chapter 10: Problem 12

Draw a parabola using the given facts. Opens right Passes through \((-2,1)\) Vertex \((-3,2)\) \(x\) -intercept \((1,0)\)

Short Answer

Expert verified
The parabola is \((y - 2)^2 = (x + 3)\).

Step by step solution

01

Understand the properties of the parabola

A parabola that opens to the right has the general equation form \[ (y - k)^2 = 4p(x - h) \] where \((h, k)\) is the vertex of the parabola. Since the vertex is given as \((-3, 2)\), we have \(h = -3\) and \(k = 2\).
02

Substitute vertex coordinates into the equation

Substituting the vertex coordinates \((-3, 2)\) into the general equation form, we get \[ (y - 2)^2 = 4p(x + 3) \] This equation reflects the vertex at \((-3, 2)\) and the direction opening to the right.
03

Use the point on the parabola to find parameter \(p\)

Since the parabola passes through \((-2,1)\), substitute \(x = -2\) and \(y = 1\) into the equation to find \(p\):\[ (1 - 2)^2 = 4p(-2 + 3) \]This simplifies to: \[ 1 = 4p \]Thus, \(p = \frac{1}{4} \).
04

Verify with the x-intercept

Check the x-intercept \((1,0)\) with the derived equation to verify the correctness:\[ (0 - 2)^2 = 4\left(\frac{1}{4}\right)(1 + 3) \]\[ 4 = 4 \] Since the equation is satisfied, \(p\) and the derived equation are 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 is crucial for identifying its shape and position in the coordinate plane. For parabolas that open horizontally, the vertex form is given by \[ (y - k)^2 = 4p(x - h) \] Here,
  • \((h, k)\) represents the vertex, which is the peak or the lowest point (depending on the direction) of the parabola.
  • \(p\) is a parameter that indicates how far and in which direction the parabola opens from the vertex.
In this exercise, the vertex is at \((-3, 2)\), which means the equation can start as \[ (y - 2)^2 = 4p(x + 3) \]. The vertex form allows us to visualize and accurately graph the parabola on the coordinate system.
Directional Opening
The direction in which a parabola opens is determined by the parameter \(p\) in its equation. For horizontally oriented parabolas, the direction is horizontal.
  • If \(p > 0\), the parabola opens to the right.
  • If \(p < 0\), the parabola opens to the left.
In the provided problem, you are informed that the parabola opens to the right, which implies a positive \(p\). Knowing this is vital as it influences how you draw or interpret the graph of the equation on the coordinate plane. The steps in the solution confirm the positive \(p\) as the values correspond to an outward opening to the right, aligning with the problem's conditions.
Point on the Parabola
A point on the parabola is used to discern additional parameters of the parabolic curve, specifically \(p\) in the vertex formula. Suppose a parabola passes through the point \((-2,1)\).
  • The coordinates \(x = -2\) and \(y = 1\) can be substituted into the equation \((y - 2)^2 = 4p(x + 3)\).
  • By substituting these, we solve for \(p\) to further define the equation.
Through the calculation: \[ (1 - 2)^2 = 4p(-2 + 3) \] This simplifies to \[ 1 = 4p \], giving \(p = \frac{1}{4} \), which uniquely tailors the parabola equation to include this specified point.
X-intercepts
The x-intercepts of a parabola are points where the graph touches or crosses the x-axis, meaning the y-coordinate is zero. Identifying x-intercepts is important as they provide notable points for graphing.In the given problem, the parabola has an x-intercept at \((1, 0)\). You can substitute this point into the equation and verify correctness, ensuring all calculations align.
  • Substitute \((1, 0)\) into the found equation: \[ (0 - 2)^2 = 4\left(\frac{1}{4}\right)(1 + 3) \]
  • This checks out, as \[ 4 = 4 \], confirming the x-intercept matches the derived curve.
This verification solidifies the solution's accuracy and assures the student they have identified a correct representation of the graph.

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Most popular questions from this chapter

Write each equation in standard form, if it is not already so, and graph it. The problems include equations that describe circles, parabolas, ellipses, and hyperbolas. $$ y=-x^{2}+6 x-4 $$

Solve each system of equations for real values of \(x\) and \(y.\) $$ \left\\{\begin{array}{l} x^{2}=4-y \\ y=x^{2}+2 \end{array}\right. $$

Solve each system of equations for real values of \(x\) and \(y.\) $$ \left\\{\begin{array}{l} 2 x^{2}-6 y^{2}+3=0 \\ 4 x^{2}+3 y^{2}=4 \end{array}\right. $$

Solve each system of equations for real values of \(x\) and \(y.\) $$ \left\\{\begin{array}{l} x^{2}-y=0 \\ x^{2}-4 x+y=0 \end{array}\right. $$

Fill in the blanks. The graph of \(2 x+y=10\) is a _____ and the graph of \(x^{2}+y^{2}=25\) is a _____.
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