Problem 14
Question
In Exercises 7 through 28 , draw a sketch of the graph of the equation. $$ x=y^{2}+1 $$
Step-by-Step Solution
Verified Answer
Plot the vertex at (1, 0) and several points using the equation, then draw a curve through them to form a parabola opening to the right.
1Step 1: Understand the Equation
The given equation is \[ x = y^2 + 1 \]. This represents a parabola. Note that it is a parabola that opens along the x-axis because the variable y is squared.
2Step 2: Identify the Vertex
The vertex of the equation \[ x = y^2 + 1 \] can be found by setting \[ y = 0 \]. This gives \[ x = 0^2 + 1 = 1 \]. Therefore, the vertex is at the point (1, 0).
3Step 3: Determine the Direction
Since the coefficient of \[ y^2 \] is positive, the parabola opens to the right along the x-axis.
4Step 4: Plot Key Points
Select some values of \[ y \] to find corresponding \[ x \] values. For example: \[ y = 1 \] gives \[ x = 1^2 + 1 = 2 \] resulting in the point (2, 1). \[ y = -1 \] gives \[ x = (-1)^2 + 1 = 2 \] resulting in the point (2, -1). Do this for a few more points: \[ y = 2 \] gives \[ x = 2^2 + 1 = 5 \] corresponding to (5, 2), and \[ y = -2 \] gives \[ x = 5 \] corresponding to (5, -2).
5Step 5: Sketch the Graph
Plot the vertex (1, 0) and the points (2, 1), (2, -1), (5, 2), and (5, -2) on a coordinate plane. Draw a smooth curve through these points to form the parabola that opens to the right.
Key Concepts
parabolasvertex of a parabolaplotting points on a graphcoordinate geometry
parabolas
A parabola is a symmetrical, curved shape that forms when graphing quadratic equations. A basic quadratic equation is typically represented as either \[ y = ax^2 + bx + c \] or \[ x = ay^2 + by + c \], where 'a,' 'b,' and 'c' are constants. The equation determines the direction and openness of the curve. If 'a' is positive, the parabola opens upward (or to the right for functions involving y); if 'a' is negative, it opens downward (or to the left).
In the given exercise, the parabola is represented by the equation \[ x = y^2 + 1 \]. Here, 'a' is 1, so the parabola opens to the right since the y term is squared. Understanding these properties helps in graphing parabolas accurately.
In the given exercise, the parabola is represented by the equation \[ x = y^2 + 1 \]. Here, 'a' is 1, so the parabola opens to the right since the y term is squared. Understanding these properties helps in graphing parabolas accurately.
vertex of a parabola
The vertex is a key point on the parabola that represents its highest or lowest point. For parabolas that open along the x-axis, like in the equation \[ x = y^2 + 1 \], the vertex can be found by setting the non-squared variable (in this case, y) to zero.
Setting \[ y = 0 \] in our equation gives: \[ x = 0^2 + 1 = 1 \].
Therefore, the vertex is at (1, 0). This point acts as a reference to help plot the rest of the graph. Knowing the vertex allows us to sketch the axis of symmetry—the imaginary line that divides the parabola into two mirror-image halves.
Setting \[ y = 0 \] in our equation gives: \[ x = 0^2 + 1 = 1 \].
Therefore, the vertex is at (1, 0). This point acts as a reference to help plot the rest of the graph. Knowing the vertex allows us to sketch the axis of symmetry—the imaginary line that divides the parabola into two mirror-image halves.
plotting points on a graph
Once you've identified the vertex, the next step is to find additional points to help outline the curve of the parabola. This is done by selecting values for 'y' and calculating the corresponding 'x' values. Here’s how:
- For \[ y = 1 \]: \[ x = 1^2 + 1 = 2 \] results in the point (2, 1).
- For \[ y = -1 \]: \[ x = (-1)^2 + 1 = 2 \] results in the point (2, -1).
- For \[ y = 2 \]: \[ x = 2^2 + 1 = 5 \] results in the point (5, 2).
- For \[ y = -2 \]: \[ x = 5 \] results in the point (5, -2).
Plot these points on a graph to form a clear shape and ensure the parabola is correctly drawn.
- For \[ y = 1 \]: \[ x = 1^2 + 1 = 2 \] results in the point (2, 1).
- For \[ y = -1 \]: \[ x = (-1)^2 + 1 = 2 \] results in the point (2, -1).
- For \[ y = 2 \]: \[ x = 2^2 + 1 = 5 \] results in the point (5, 2).
- For \[ y = -2 \]: \[ x = 5 \] results in the point (5, -2).
Plot these points on a graph to form a clear shape and ensure the parabola is correctly drawn.
coordinate geometry
Coordinate geometry, or analytic geometry, is the study of geometry using a coordinate system. This system helps in solving geometric problems by placing shapes in a more calculated and graphical form. Key elements include:
- **Coordinate plane**: A two-dimensional plane formed by the x-axis (horizontal) and y-axis (vertical), meeting at the origin (0,0).
- **Points**: Represented as (x, y), points are locations on the coordinate plane.
In the context of this exercise, each calculated point (like (2, 1) or (5, -2)) is plotted on the coordinate plane to sketch the parabola. Through coordinate geometry, we can visualize and solve equations systematically, making abstract algebraic concepts easier to understand.
- **Coordinate plane**: A two-dimensional plane formed by the x-axis (horizontal) and y-axis (vertical), meeting at the origin (0,0).
- **Points**: Represented as (x, y), points are locations on the coordinate plane.
In the context of this exercise, each calculated point (like (2, 1) or (5, -2)) is plotted on the coordinate plane to sketch the parabola. Through coordinate geometry, we can visualize and solve equations systematically, making abstract algebraic concepts easier to understand.
Other exercises in this chapter
Problem 14
In Exercises 11 through 14 , find the center and radius of each circle, and draw a sketch of the graph. $$ x^{2}+y^{2}-10 x-10 y+25=0 $$
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In Exercises 11 through 32 , find the solution set of the given inequality and illustrate the solution on the real number $$ 3 x-5
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Find an equation that must be satisfied by the coordinates of any point that is equidistant from the two points \((-3,2)\) and \((4,6)\).
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In Exercises 5 through 14, find an equation of the line satisfying the given conditions. $$ \text { The slope is }-2 \text {, and the } x \text { intercept is }
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