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Chapter 3: Problem 60

Write the quadratic function in vertex form. Then identify the vertex. \(f(x)=x^2+6 x-16\)

Short Answer

Expert verified
Vertex form: \(f(x)= (x+3)^2-25\). Vertex is at (-3, -25).

Step by step solution

01

Write the function

Start by writing down the given quadratic function \(f(x)=x^2+6 x-16\)
02

Group the x terms

Group the terms containing x together: \(f(x)= (x^2+6x) -16\)
03

Complete the square

In order to complete the square, find a number that when added and subtracted in the parentheses will make it a perfect square trinomial. We need to add and subtract \((b/2)^2\), where b is the coefficient of x. In our function, b=6. So, \( (b/2)^2=(6/2)^2=9\). Therefore, the equation becomes: \(f(x)=(x^2+6x+9-9) -16 = (x^2+6x+9)-9 -16 = (x+3)^2-25\)
04

Identify the vertex

Now we can see the equation in vertex form \(f(x)=a(x-h)^2 + k\) where a=1, h=-3 and k=-25. So, the vertex of the parabola is at point (-3,-25)

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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 an essential concept in algebra, commonly expressed as a polynomial of degree two. These functions are defined by the standard form equation \[ f(x) = ax^2 + bx + c \]where:
  • \( a \) is the leading coefficient, which cannot be zero,
  • \( b \) is the coefficient of \( x \), and
  • \( c \) is the constant term.
The graph of a quadratic function is a parabola, which is a symmetrical curve that either opens upwards or downwards.
Whether it opens upward or downward depends on the sign of \( a \):
  • If \( a > 0 \), the parabola opens upward.
  • If \( a < 0 \), the parabola opens downward.
Understanding quadratic functions is crucial as they are widely used in various fields including physics, engineering, and finance due to their ability to model real-world situations like projectile motion and profit optimization.
Completing the Square
Completing the square is a technique used to transform a quadratic equation into its vertex form. This process makes it easier to identify features of the parabola such as its vertex. To complete the square:
  • Start with a quadratic function, for example, \( f(x) = x^2 + 6x - 16 \).
  • Group the terms that contain \( x \) together, separating them from the constant: \( f(x) = (x^2 + 6x)-16 \).
  • Calculate the term to add and subtract to make a perfect square trinomial. This term is calculated as \( (b/2)^2 \). Here \( b = 6 \), so \( (6/2)^2 = 9 \).
  • Form the perfect square by adjusting the equation: \( f(x) = (x^2 + 6x + 9 - 9) -16 = (x+3)^2 - 25 \).
By completing the square, the quadratic can be rewritten in vertex form:\[ f(x) = a(x-h)^2 + k \]This form simplifies understanding the parabola's graph.
Vertex of a Parabola
The vertex of a parabola is a significant point that represents either the maximum or minimum of a quadratic function, depending on the parabola's orientation. In vertex form, \[ f(x) = a(x-h)^2 + k \], the vertex can be easily identified:
  • \( h \) is the x-coordinate of the vertex,
  • \( k \) is the y-coordinate of the vertex.
For the completed example \( f(x) = (x+3)^2 - 25 \), we can directly read off the vertex:
  • The x-coordinate \( h = -3 \) (note the negative sign used in the vertex form),
  • The y-coordinate \( k = -25 \).
Thus, the vertex is at the point \((-3, -25)\), and since \( a = 1 \) which is positive, the parabola opens upwards indicating that the vertex is its lowest point.

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