Problem 18
Question
The graph of each equation is a parabola. Find the vertex of the parabola and then graph it. $$x=y^{2}-6 y+6$$
Step-by-Step Solution
Verified Answer
The vertex of the parabola is \((-3, 3)\). The parabola opens to the right.
1Step 1: Reorganize the Equation
The given equation is \( x = y^2 - 6y + 6 \). We want to express this in a form that makes it easy to identify the vertex. Our aim is to complete the square for the \( y \) terms.
2Step 2: Complete the Square
To complete the square, we consider the \( y^2 - 6y \) portion. Take half of the \( y \) coefficient, which is -6, divide it by 2 to get -3, then square it to get 9. Add and subtract 9 inside the equation: \( x = (y^2 - 6y + 9) - 9 + 6 \).
3Step 3: Rewrite the Equation
Now, rewrite the equation with the perfect square: \( x = (y - 3)^2 - 3 \). This shows the equation of the parabola in vertex form.
4Step 4: Identify the Vertex
The vertex form of a parabola \((y - k)^2 = 4p(x - h)\) gives the vertex at \((h, k)\). Here, \((y - 3)^2 = x + 3\), and the vertex is thus \((-3, 3)\).
5Step 5: Graph the Parabola
Plot the vertex \((-3, 3)\) on a coordinate plane. Since the coefficient of the squared term \((y - 3)^2\) is positive, the parabola opens to the right. The equation can be rewritten as \(x + 3 = (y - 3)^2\), indicating the shift by -3 along the x-axis.
Key Concepts
Completing the SquareVertex Form of a ParabolaCoordinate Plane Graphing
Completing the Square
Completing the square is a fundamental technique that helps transform quadratic expressions into a more manageable form. Let's dive into why it's useful and how it works. For a quadratic expression like \( y^2 - 6y \), completing the square involves converting this into a perfect square trinomial. Here's how:
- Take the coefficient of the linear term, which in this case is -6.
- Divide this coefficient by 2 to get -3.
- Square the result, obtaining \((-3)^2 = 9\).
Vertex Form of a Parabola
The vertex form of a parabola is particularly advantageous when you need to quickly identify important properties of the graph, such as the vertex. A parabola expressed in vertex form looks like \((y - k)^2 = 4p(x - h)\) for horizontal parabolas, where \((h, k)\) represents the vertex.
In the exercise example, after completing the square, the equation becomes \(x = (y - 3)^2 - 3\). This reveals the parabola's vertex, \((-3, 3)\).
In the exercise example, after completing the square, the equation becomes \(x = (y - 3)^2 - 3\). This reveals the parabola's vertex, \((-3, 3)\).
- The vertex form directly tells you the point which is essentially the 'tip' or 'turning point' of the parabola.
- Understanding the structure of this form helps in predicting the direction the parabola opens – to the right or left for horizontal parabolas.
Coordinate Plane Graphing
Graphing on the coordinate plane involves plotting points and understanding how algebraic expressions translate into visual representations. With simpler forms like the vertex form, we can efficiently plot parabolas by finding key features such as the vertex and direction of opening.
Here's how you address graphing using the example equation \(x = (y - 3)^2 - 3\):
Here's how you address graphing using the example equation \(x = (y - 3)^2 - 3\):
- First, take the vertex \((-3, 3)\) and plot it on the graph.
- Since the equation shows a positive coefficient for the squared term, \((y - 3)^2\), the parabola opens to the right.
- Consider additional points by choosing values for \(y\) and solving for \(x\) to complete the parabola's shape.
Other exercises in this chapter
Problem 18
Graph each inequality. See Examples 1 and \(2 .\) $$ y>(x-2)^{2}+1 $$
View solution Problem 18
Sketch the graph of each equation. \(\frac{(x+3)^{2}}{16}+\frac{(y+2)^{2}}{4}=1\)
View solution Problem 19
Solve each nonlinear system of equations for real solutions. $$ \left\\{\begin{aligned} 2 x^{2}+3 y^{2} &=14 \\ -x^{2}+y^{2} &=3 \end{aligned}\right. $$
View solution Problem 19
Sketch the graph of each equation. \(\frac{x^{2}}{4}-\frac{y^{2}}{9}=1\)
View solution