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Chapter 8: Problem 31

Find the vertex for the graph of each quadratic function. $$y=x^{2}-4 x+1$$

Short Answer

Expert verified
The vertex of the quadratic function is (2, -3).

Step by step solution

01

Identify the quadratic function form

The given quadratic function is of the form \(y = ax^2 + bx + c\). Here, the coefficients are:\(a = 1\), \(b = -4\), and \(c = 1\).
02

Use the vertex formula for x-coordinate

For a quadratic function in the form \(y = ax^2 + bx + c\), the x-coordinate of the vertex can be found using the formula: \(x = -\frac{b}{2a}\).Substitute the values of \(a\) and \(b\) into the formula:\(x = -\frac{-4}{2(1)} = \frac{4}{2} = 2\).
03

Find the y-coordinate of the vertex

To find the y-coordinate, substitute the x-coordinate of the vertex back into the original quadratic equation. Given \(x = 2\), substitute it into \(y = x^2 - 4x + 1\):\(y = (2)^2 - 4(2) + 1 = 4 - 8 + 1 = -3\).

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Key Concepts

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

quadratic equations
Quadratic equations form the cornerstone of many algebraic concepts and have a general form expressed as:
\(y = ax^2 + bx + c\).
These equations always graph into a U-shaped curve known as a parabola.
In a quadratic equation, the highest exponent of the variable, usually \(x\), is 2. This makes them different from linear equations, where the highest exponent is 1.
Quadratic equations can be solved using various methods such as:
  • Factoring
  • Completing the square
  • Using the quadratic formula
vertex formula
The vertex of a quadratic function is a key component. It is the highest or lowest point on the graph of the function, depending on the parabola's orientation.
To find the vertex, we use the vertex formula:
\( x = -\frac{b}{2a} \).
Here, \(x\) is the x-coordinate of the vertex, and \(a\) and \(b\) are coefficients from the quadratic equation \(y = ax^2 + bx + c\).
Once we find the x-coordinate, we can substitute it back into the original equation to find the y-coordinate.
This way, you can precisely locate the vertex of your quadratic function.
parabola properties
Parabolas are unique curves derived from quadratic equations.
Several properties define a parabola, including:
  • Vertex: The peak or the lowest point on the graph.
  • Axis of symmetry: A vertical line that passes through the vertex and divides the parabola into two symmetrical halves.
  • Direction: If \(a\) is positive, the parabola opens upwards. If \(a\) is negative, it opens downwards.
  • Focus and Directrix: Special lines associated with the parabolic curve, which help in geometric interpretations.
Understanding these properties is crucial for drawing and analyzing quadratic functions.
standard form of quadratic function
The standard form of a quadratic function is written as:
\(y = ax^2 + bx + c\).
Here, \(a\), \(b\), and \(c\) are constants.
This form allows for easy application of the vertex formula and other methods of solving quadratic equations.
For example, in the equation \(y = x^2 - 4x + 1\), we identified:
  • \(a = 1\)
  • \(b = -4\)
  • \(c = 1\)
Using these coefficients, we determined the vertex to understand the graph's peak accurately. The standard form is a foundational aspect of working with quadratic equations.

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Most popular questions from this chapter

Find all real or imaginary solutions to each equation. Use the method of your choice. $$\frac{1}{x}+\frac{1}{x-1}=\frac{1}{4}$$

Determine the number of real solutions to each equation by examining the calculator graph of \(y=a x^{2}+b x+c .\) Use the discriminant to check your conclusions. $$-2 x^{2}-403=0$$

Use the quadratic formula and a calculator to solve each equation. Round answers to three decimal places and check your answers. $$x^{2}+3.2 x-5.7=0$$

Find all real or imaginary solutions to each equation. Use the method of your choice. $$4 x^{2}+25=0$$

What is the product of the two solutions to \(6 x^{2}+5 x-4=0 ?\) Explain why the product of the solutions to any quadratic equation is \(\frac{c}{a}\)
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