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Chapter 3: Problem 64

Derive the formula \(x=-b\) for the \(x\) -coordinate of the vertex of parabola \(y=a(x+b)^{2}+c .\)

Short Answer

Expert verified
The x-coordinate of the vertex is \(x = -b\).

Step by step solution

01

Recall the Vertex Form of a Parabola

The general vertex form of a parabola is given by \(y = a(x-h)^2 + k\), where \((h, k)\) is the vertex of the parabola. Here, \(h\) is the x-coordinate of the vertex.
02

Match the Given Equation with the Vertex Form

The given equation is \(y = a(x + b)^2 + c\). This can be rewritten to match the vertex form: \(y = a(x - (-b))^2 + c\). In this representation, it is clear that \(-b\) is taking the place of \(h\) from the vertex form.
03

Identify the Vertex x-Coordinate

From our rewritten equation \(y = a(x - (-b))^2 + c\), we see that the x-coordinate of the vertex should be \(x = -b\). Here, we have determined that \(h = -b\).
04

Conclusion

Thus, we have derived that the x-coordinate of the vertex of the given parabola \(y = a(x+b)^2+c\) is \(x = -b\).

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Key Concepts

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

Quadratic Functions
Quadratic functions form the foundational elements of algebra when it comes to analyzing parabolas. A quadratic function is typically expressed as \( y = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. These functions can create a U-shaped curve on a graph, known as a parabola.

Here's what you should know about quadratic functions:
  • The leading coefficient \( a \) determines the direction of the parabola. If \( a > 0 \), the parabola opens upwards. Conversely, if \( a < 0 \), it opens downwards.
  • The vertex of the parabola is the highest or lowest point, depending on the direction of the parabola.
  • The axis of symmetry is a vertical line that passes through the vertex, dividing the parabola into two mirror-image halves.
These properties make quadratic functions uniquely interesting and useful in various real-world applications, such as projectile motion and biological models.
Vertex Form
Converting a quadratic equation into vertex form helps in easily identifying the vertex of the parabola. The vertex form of a quadratic function is \( y = a(x-h)^2 + k \), where \( (h, k) \) is the vertex of the parabola.

Understanding the vertex form provides several benefits:
  • It directly gives you the coordinates of the vertex, \( (h, k) \), without additional calculations.
  • The \( a \) in the vertex form still dictates the opening direction and width of the parabola, just as it does in the standard form.
  • Translating between standard form and vertex form involves completing the square, a key algebraic technique.
Having the equation in vertex form makes graphing the parabola straightforward, as you can easily pinpoint its central features like the direction, width, and the vertex.
Parabolic Equations
Parabolic equations define the paths of parabolas and hold significant importance in various fields. These equations come in several forms, but they all describe shapes that are symmetric and have a distinct vertex.

Some notable characteristics of parabolic equations include:
  • The vertex form \( y = a(x-h)^2 + k \) explicitly reveals the vertex and is preferred for graphical analysis.
  • The standard form \( y = ax^2 + bx + c \) is often easier for solving algebraically due to its straightforward structure.
  • Both forms ultimately model the same parabolic curve, although the manipulation required to derive each from one another highlights different aspects of the parabola.
Parabolic equations are more than just shapes鈥攖hey represent the predictable and constant acceleration in physics, the elliptical orbits in astronomy, and much more. Understanding these equations aids in solving practical problems across numerous scientific disciplines.

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Most popular questions from this chapter

In general, implicit differentiation gives an expression for the derivative that involves both \(x\) and \(y\). Under what conditions will the expression involve only \(x ?\)

GENERAL: Airplane Flight Path A plane is to take off and reach a level cruising altitude of 5 miles after a horizontal distance of 100 miles, as shown in the following diagram. Find a polynomial flight path of the form \(f(x)=a x^{3}+b x^{2}+c x+d\) by following Steps i to iv to determine the constants \(a, b, c,\) and \(d\). i. Use the fact that the plane is on the ground at \(x=\) 0 [that is, \(f(0)=0]\) to determine the value of \(d\). ii. Use the fact that the path is horizontal at \(x=0\) [that is, \(\left.f^{\prime}(0)=0\right]\) to determine the value of \(c\). iii. Use the fact that at \(x=100\) the height is 5 and the path is horizontal to determine the values of \(a\) and \(b\). State the function \(f(x)\) that you have determined. iv. Use a graphing calculator to graph your function on the window [0,100] by [0,6] to verify its shape.

In each equation, \(x\) and \(y\) are functions of \(t\) Differentiate with respect to \(t\) to find a relation between \(d x / d t\) and \(d y / d t\). $$ x^{2} y=80 $$

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Given a Cobb-Douglas production relation $$P=a L^{b} K^{1-b} \quad a>0 \text { and } 0 < b < 1$$ giving the total production \(P\) from \(L\) units of labor and \(K\) units of capital, along any isoquant (that is, for a fixed production level), \(\frac{d K}{d L}\) may be found by implicit differentiation. The absolute value of this derivative is called the marginal rate of technical substitution of labor for capital and is denoted MRTS. Show that \(M R S T=\frac{b}{1-b} \frac{K}{L}\) along any given isoquant.
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