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Chapter 10: Problem 12

The graph of each equation is a parabola. Find the vertex of the parabola and then graph it. $$y=-2 x^{2}$$

Short Answer

Expert verified
The vertex of the parabola is at (0, 0).

Step by step solution

01

Identify the General Form of the Equation

The given equation is \(y = -2x^2\). Identify the general form of a parabola equation as \(y = ax^2 + bx + c\). Here, \(a = -2\), \(b = 0\), and \(c = 0\).
02

Find the Vertex Formula

The vertex of a parabola given by \(y = ax^2 + bx + c\) is found using the formula \(x = -\frac{b}{2a}\). As \(b = 0\), the formula simplifies to \(x = 0\).
03

Calculate y-coordinate of the Vertex

Substitute \(x = 0\) into the equation \(y = -2x^2\). Thus, \(y = -2(0)^2 = 0\). So, the vertex's \(y\)-coordinate is 0.
04

Determine the Vertex of the Parabola

Combine the results of the \(x\)- and \(y\)-coordinates: The vertex is \((0, 0)\).
05

Graph the Parabola

Plot the vertex \((0, 0)\) on the coordinate plane. The parabola opens downwards because \(a = -2\) is negative. Sketch the downward curve passing through the vertex.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Graphing Parabolas
Graphing parabolas is a fundamental skill in working with quadratic equations, as these create the characteristic U-shape curve on a graph. A parabola can open upwards or downwards, and the direction it opens depends on the coefficient (\(a\)) in the quadratic equation of the form \(y = ax^2 + bx + c\).

Here's how you graph a parabola:
  • Find the vertex, which is the highest or lowest point on the graph, depending on the parabola's orientation.
  • Determine the direction it opens: if \(a > 0\), the parabola opens upwards; if \(a < 0\), it opens downwards.
  • Plot additional points on either side of the vertex to understand the shape better.
For the equation given, \(y = -2x^2\), the parabola opens downwards since the \(a\) value is \(-2\), which is less than zero. By plotting the vertex at (0,0) and sketching the curve, you can visualize how it extends in both directions, creating a symmetrical shape about the y-axis.
Quadratic Equations
Quadratic equations often appear in the form \(y = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants. This type of equation describes a parabola, which is a symmetrical curve on a graph. Understanding the components of this equation is crucial for interpreting and solving various problems.

Each term in the equation serves a purpose:
  • \(a\): Determines the direction and the "width" of the parabola. If \(a\) is positive, the parabola opens upwards; if negative, it opens downwards.
  • \(b\): Influences the position of the vertex along the x-axis, although it is zero in the provided equation, simplifying the process.
  • \(c\): Gives the y-intercept, showing where the parabola crosses the y-axis.
In our example, \(y = -2x^2\), both \(b\) and \(c\) are zero, focusing our attention on the fact that the vertex is located at the origin, (0,0).
Vertex Formula
The vertex of a parabola is a significant point that provides essential information about the parabola's position and shape. The vertex formula helps in locating this point, especially for quadratic equations. To find the vertex, you use the formula \(x = -\frac{b}{2a}\), which gives you the x-coordinate.

For example, consider the equation \(y = -2x^2\):
  • \(b = 0\), so when you plug it into the vertex formula, it simplifies to \(x = 0\).
  • To find the y-coordinate for the vertex, substitute \(x = 0\) back into the original equation to get \(y = -2(0)^2 = 0\).
Hence, the vertex is \((0, 0)\). This information is vital for graphing and understanding the parabola's nature, as it represents the peak or the lowest point depending on whether the parabola opens down or up.

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Most popular questions from this chapter

Find each function value if \(f(x)=3 x^{2}-2 .\) See Section 3.2 $$ f(a) $$

The orbits of stars, planets, comets, asteroids, and satellites all have the shape of one of the conic sections. Astronomers use a measure called eccentricity to describe the shape and elongation of an orbital path. For the circle and ellipse, eccentricity e is calculated with the formula \(e=\frac{c}{d},\) where \(c^{2}=\left|a^{2}-b^{2}\right|\) and \(d\) is the larger value of a or b. For a hyperbola, eccentricity e is calculated with the formula \(e=\frac{c}{d},\) where \(c^{2}=a^{2}+b^{2}\) and the value of \(d\) is equal to a if the hyperbola has \(x\) -intercepts or equal to b if the hyperbola has \(y\) -intercepts. A. \(\frac{x^{2}}{36}-\frac{y^{2}}{13}=1\) B. \(\frac{x^{2}}{4}+\frac{y^{2}}{4}=1\) C. \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\) D. \(\frac{y^{2}}{25}-\frac{x^{2}}{39}=1\) G. \(\frac{x^{2}}{16}-\frac{y^{2}}{65}=1\) E. \(\frac{x^{2}}{17}+\frac{y^{2}}{81}=1\) F. \(\frac{x^{2}}{36}+\frac{y^{2}}{36}=1\) H. \(\frac{x^{2}}{144}+\frac{y^{2}}{140}=1\) For each of the equations \(A-H,\) identify the values of \(a^{2}\) and \(b^{2}\).

For the exercises below, see the Concept Check in this section. Without solving, how can you tell that the graphs of \(y=2 x+3\) and \(y=2 x+7\) do not have any points of intersection?

Graph each inequality in two variables. $$ 3 x-y \leq 4 $$

Solve each nonlinear system of equations for real solutions. $$ \left\\{\begin{aligned} x^{2}+y^{2} &=25 \\ 4 x+3 y &=0 \end{aligned}\right. $$
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