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Chapter 3: Problem 25

Find the vertex of the parabola by applying the vertex formula. $$ k(a)=-\frac{1}{3} a^{2}+6 a+1 $$

Short Answer

Expert verified
(9, 28)

Step by step solution

01

Identify coefficients

In the given quadratic function, identify the coefficients for the terms. The general form of a quadratic function is \[f(a)=ax^2+bx+c\]. Here, the function is \[k(a)=-\frac{1}{3}a^2+6a+1\]. Therefore, \ a=-\frac{1}{3} \, \ b=6 \, and \ c=1 \.
02

Apply the vertex formula

The vertex formula for finding the x-coordinate of the vertex for the parabola given by \[f(x) = ax^2 + bx + c\] is \[x = \frac{-b}{2a}\]. Substitute \ a \, and \ b \ into the formula: \[x=\frac{-6}{2(-\frac{1}{3})}\].
03

Simplify the x-coordinate

Calculate the value by simplifying the expression \[x = \frac{-6}{2(-\frac{1}{3})} = \frac{-6}{-\frac{2}{3}} = \frac{-6 \cdot -3}{2} = 9\]. So, the x-coordinate of the vertex is \ a = 9 \.
04

Find the y-coordinate

Substitute \ a = 9 \ back into the original quadratic function \[k(a) = -\frac{1}{3} a^2 + 6a + 1 \] to find the y-coordinate. \[ k(9) = -\frac{1}{3} (9)^2 + 6(9) + 1 \]. Simplify: \[ k(9) = -\frac{1}{3} (81) + 54 + 1 \]. \[ = -27 + 54 + 1 = 28 \]. This gives \ y = 28 \.
05

State the vertex

Thus, the vertex of the parabola \[k(a) = -\frac{1}{3} a^2 + 6a + 1 \] is at the point \ (9, 28) \.

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Key Concepts

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

Quadratic Functions
A quadratic function is an equation of the form
\[ f(x) = ax^2 + bx + c \].
This type of function graphs as a curve called a parabola.
The highest or lowest point of a parabola, known as the vertex,
occurs where the function changes direction.
Quadratic functions can open upwards, making a U-shape,
or downwards, creating an inverted U-shape.
This behavior depends on the sign of the coefficient \(a\):
  • If \(a > 0\), the parabola opens upwards.
  • If \(a < 0\), it opens downwards.
Vertex Formula
To find the vertex of a quadratic function, we use the vertex formula for the x-coordinate:
\[ x = \frac{-b}{2a} \].
This formula comes from completing the square or using calculus.
After calculating the x-coordinate, you substitute it back into the original quadratic equation to find the y-coordinate.
In our example, let's break it down step by step:
  • For \[ k(a) = -\frac{1}{3} a**2 + 6a + 1 \], we identified \(a = -\frac{1}{3}\)
  • Using the x-coordinate formula: \[ x = \frac{-6}{2(-\frac{1}{3})} \], Simplifying, we get \(x = 9\)
  • Then, to find the y-coordinate, substitute \(a = 9\) into the equation: \[ k(9) = -\frac{1}{3} (9)^2 + 6(9) + 1 \]
  • Simplify to get \( k(9) = 28 \), making the coordinates of the vertex \((9, 28)\)
Parabola
A parabola is a U-shaped curve that represents the graph of a quadratic function.
The vertex of the parabola is the point where it reaches its maximum or minimum value.
Understanding the movement and shape of parabolas help in many practical applications:
  • Projectile motion
  • Optics
  • Economics for profit maximization
Here are a few key characteristics of parabolas:
  • They are symmetric around a vertical line called the axis of symmetry.
  • The direction is determined by the sign of the leading coefficient.
By applying methods like the vertex formula, you can easily identify significant points on the graph,
which helps in analyzing and interpreting quadratic functions.

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Most popular questions from this chapter

The light from a lightbulb radiates outward in all directions. a. Consider the interior of an imaginary sphere on which the light shines. The surface area of the sphere is directly proportional to the square of the radius. If the surface area of a sphere with a \(10-\mathrm{m}\) radius is \(400 \pi \mathrm{m}^{2}\), determine the surface area of a sphere with a \(20-\mathrm{m}\) radius. b. Explain how the surface area changed when the radius of the sphere increased from \(10 \mathrm{~m}\) to \(20 \mathrm{~m}\). c. Based on your answer from part (b) how would you expect the intensity of light to change from a point \(10 \mathrm{~m}\) from the lightbulb to a point \(20 \mathrm{~m}\) from the lightbulb? d. The intensity of light from a light source varies inversely as the square of the distance from the source. If the intensity of a lightbulb is 200 lumen/m \(^{2}\) (lux) at a distance of \(10 \mathrm{~m}\), determine the intensity at \(20 \mathrm{~m}\).

A power company burns coal to generate electricity. The cost \(C(x)\) (in \(\$ 1000\) ) to remove \(x \%\) of the air pollutants is given by $$C(x)=\frac{600 x}{100-x}$$ a. Compute the cost to remove \(25 \%\) of the air pollutants. $$(\text { Hint }: x=25 .)$$ b. Determine the cost to remove \(50 \%, 75 \%,\) and \(90 \%\) of the air pollutants. c. If the power company budgets \(\$ 1.4\) million for pollution control, what percentage of the air pollutants can be removed?

A food company originally sells cereal in boxes with dimensions 10 in. by 7 in. by 2.5 in. To make more profit, the company decreases each dimension of the box by \(x\) inches but keeps the price the same. If the new volume is \(81 \mathrm{in} .{ }^{3}\) by how much was each dimension decreased?

Write a statement in words that describes the variation model given. Use \(k\) as the constant of variation. \(E=\frac{k c^{2}}{\sqrt{b}}\)

Determine if the statement is true or false. If 5 is an upper bound for the real zeros of \(f(x)\), then 4 is also an upper bound.
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