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Chapter 1: Problem 12

State whether the graph of the function is a parabola. If the graph is a parabola, then find the parabola's vertex. \(y=3 x^{2}-6 x\)

Short Answer

Expert verified
The graph is a parabola with vertex \( (1, -3) \).

Step by step solution

01

- Identify the type of function

The given function is \(y = 3x^2 - 6x\). Notice the highest power of \(x\) is 2, which means this is a quadratic function. The graph of a quadratic function is always a parabola.
02

- Write the function in vertex form

To find the vertex of the parabola, we first need to rewrite the function in the vertex form, \(y = a(x-h)^2 + k\). We start by completing the square for the expression \(3x^2 - 6x\).
03

- Complete the square

First factor out the coefficient of \(x^2\) from the terms involving \(x\). \[ y = 3(x^2 - 2x) \] To complete the square, add and subtract \((\frac{2}{2})^2 = 1\) inside the parentheses: \[ y = 3(x^2 - 2x + 1 - 1) \] Group the perfect square trinomial and simplify: \[ y = 3((x-1)^2 - 1) \] Distribute the 3: \[ y = 3(x-1)^2 - 3 \] Now the function is in vertex form, \(y = 3(x-1)^2 - 3\).
04

- Identify the vertex

In the vertex form \(y = 3(x-1)^2 - 3\), the vertex \((h, k)\) can be directly read from the equation. Here, \(h = 1\) and \(k = -3\). Therefore, the vertex of the parabola is at \((1, -3)\).

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Key Concepts

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

quadratic function
A quadratic function is any function that can be written in the form: \[ y = ax^2 + bx + c \]The highest degree term is \(x^2\), which makes it quadratic. These functions generate graphs that are parabolas. Notice the function given in the exercise is: \[ y = 3x^2 - 6x \]Here, \( a = 3 \), \( b = -6 \), and \( c = 0 \).This confirms it's a quadratic function and the graph will be a parabola. Quadratic functions have several important properties, such as:
  • Symmetry: The graph is symmetric about a vertical line called the axis of symmetry.
  • Parabolic shape: The graph is U-shaped, opening either up or down depending on the sign of a.
  • Vertex: The highest or lowest point on the parabola, depending on whether it opens down or up, respectively.
Understanding these properties helps in graphing and analyzing quadratic functions.
completing the square
Completing the square is a method used to transform a quadratic function into vertex form. This makes it easier to identify the vertex of the parabola.Let's break down the process:
  • Step 1: Start with the quadratic expression \(3x^2 - 6x\).
  • Step 2: Factor out the coefficient of \(x^2\) from terms involving \(x\): \(3(x^2 - 2x)\).
  • Step 3: Adjust within the parentheses to create a perfect square trinomial. Add and subtract \((\frac{2}{2})^2 = 1\): \(3(x^2 - 2x + 1 - 1)\).
  • Step 4: Group the perfect square trinomial \((x - 1)^2\) and simplify: \(3((x-1)^2 - 1)\).
  • Step 5: Distribute the 3 to get the final vertex form: \(3(x-1)^2 - 3\).
This approach converts the original quadratic function into a more useful form for identifying its vertex.
vertex form
The vertex form of a quadratic function is expressed as: \[ y = a(x-h)^2 + k \]Here, \(h, k\) are the coordinates of the vertex of the parabola. It shows directly how the parabola is shifted from the origin. In our example, converting \(y = 3x^2 - 6x\) to vertex form gave us:\[ y = 3(x-1)^2 - 3 \]From this vertex form, we identify the vertex as \((1, -3)\). This makes it very easy to graph the parabola, as you now know:
  • The vertex is at \((1, -3)\).
  • The axis of symmetry is the vertical line \(x = 1\).
  • The parabola opens upwards because the coefficient of \((x-1)^2\) is positive (i.e., \(a = 3\)).
Using the vertex form, plotting or analyzing the behavior of quadratic functions becomes straightforward. This form highlights how the function behaves around its vertex, which is crucial for understanding its properties.

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Most popular questions from this chapter

Solar Radiation. The annual radiation (in megajoules per square centimeter) for certain land areas of the northern hemisphere may be modeled with the equation \({ }^{19}\) \(R=0.339+0.808 \cos l \cos s-0.196 \sin l \sin s\) \(-0.482 \cos a \sin s\) In this equation, \(l\) is the latitude (between \(30^{\circ}\) and \(60^{\circ}\) ) and \(s\) is the slope of the ground (between \(0^{\circ}\) and \(60^{\circ}\) ). Also, \(a\) is the aspect, or the direction that the slope faces. For a slope facing due north, \(a=0^{\circ}\), and for a slope facing south, \(a=180^{\circ} .\) For a slope facing either east or west, \(a=90^{\circ}\). Find the annual radiation of south-facing land at \(30^{\circ}\) north latitude with a \(20^{\circ}\) slope.

Sketch the following angles. \(-\boldsymbol{\pi}\)

Use a calculator to evaluate the following trigonometric functions. $$ \tan (-2.3 \pi) $$

Find all solutions of the given equation. $$ 2 \sin \left(t+\frac{\pi}{3}\right)=-\sqrt{3} $$

Use a calculator to evaluate the following trigonometric functions. $$ \sin (1.2 \pi) $$
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