Problem 27
Question
\(23-28\) Use a graphing device to graph the parabola. $$ 4 x+y^{2}=0 $$
Step-by-Step Solution
Verified Answer
The parabola opens to the left with vertex at (0,0).
1Step 1: Rewrite the Equation in Terms of y
Start by rewriting the equation \(4x + y^2 = 0\) in a standard parabola form, which involves expressing \(y^2\) in terms of \(x\). To do this, isolate \(y^2\): Subtract \(4x\) from both sides: \[ y^2 = -4x \] This shows the parabola is oriented along the x-axis as \(y\) is squared, opening to the left since the coefficient of \(x\) is negative.
2Step 2: Identify the Parabola's Vertex and Orientation
With the equation \(y^2 = -4x\), we identify that it is of the form \((y-k)^2 = -4p(x-h)\) where \(h=0\), \(k=0\), and \(p=1\). The vertex is at \((0,0)\) and it opens to the left since the coefficient of \(x\) is negative.
3Step 3: Choose Values for x and Solve for y
Select a few x-values to find corresponding y-values and sketch the parabola. For instance, if \(x = 0\), then \(y^2 = 0\), so \(y = 0\). If \(x = -1\), then \(y^2 = 4\), giving \(y = 2\) or \(y = -2\). If \(x = -4\), then \(y^2 = 16\), so \(y = 4\) or \(y = -4\).
4Step 4: Plot the Points
Plot the calculated points \((0,0)\), \((-1,2)\), \((-1,-2)\), \((-4,4)\), \((-4,-4)\) on a graph. These points illustrate the shape and direction of the parabola.
5Step 5: Draw the Parabola
Use a smooth curve to connect the plotted points, making sure the curve opens to the left from the vertex at \((0,0)\) and that it extends infinitely in the negative x-direction.
6Step 6: Verify with a Graphing Device
Confirm the shape and orientation of the parabola using a graphing device. Input the equation \(4x + y^2 = 0\) or \(y^2 = -4x\) to check that the graph matches the parabolic curve you've sketched manually.
Key Concepts
Parabola Standard FormVertex of a ParabolaOrientation of ParabolasGraphing Techniques
Parabola Standard Form
The standard form of a parabola makes it much simpler to analyze and graph. In general, the equation of a parabola can be expressed in two standard forms:
- For parabolas opening vertically: \(y = ax^2 + bx + c\)
- For parabolas opening horizontally: \(x = ay^2 + by + c\)
Vertex of a Parabola
The vertex of a parabola is a crucial point, as it serves as the "turning point" or the point where the parabola switches direction. For the equation \(y^2 = -4x\), which is in the form \((y - k)^2 = 4p(x - h)\), the vertex \( (h, k) \) can be directly identified. Here, both \(h\) and \(k\) are zero, so the vertex of the parabola is \( (0, 0) \). Understanding the vertex's location helps in plotting the parabola correctly on a graph. It is the central point that dictates the symmetry of the parabola as it curves from one side to the other.
Orientation of Parabolas
Measuring the orientation of parabolas is essential for graphing them accurately. In our equation \(y^2 = -4x\), the squared term is on the side of \(y\), meaning the parabola opens sideways rather than up or down.
- If \(y^2 = 4px\) (positive coefficient), the parabola opens to the right.
- If \(y^2 = -4px\) (negative coefficient), the parabola opens to the left.
Graphing Techniques
Graphing a parabola manually can be facilitated with several techniques. First, it is important to identify a few key points by substituting different \(x\)-values to solve for \(y\)-values. This provides coordinates you can plot to outline the parabola's path.
- Selecting Points: Choose strategic \(x\)-values, such as \(-1\) and \(-4\), to find \(y\)-values.
- Plotting Points: Carefully plot these points on your graph to start shaping the curve.
- Drawing the Curve: Connect the points smoothly to form the parabolic shape, ensuring it reflects the correct orientation.
Other exercises in this chapter
Problem 27
\(23-34\) Complete the square to determine whether the equation represents an ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an ellips
View solution Problem 27
\(15-28=(a)\) Use the discriminant to determine whether the graph of the equation is a parabola, an ellipse, or a hyperbola. (b) Use a rotation of axes to elimi
View solution Problem 28
Use a graphing device to graph the hyperbola. $$ 3 y^{2}-4 x^{2}=24 $$
View solution Problem 28
A polar equation of a conic is given. (a) Show that the conic is a hyperbola, and sketch its graph. (b) Find the vertices anddirectrix, and indicate them on the
View solution