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Chapter 2: Problem 18

\(y=(x-2)^2-1\)

Short Answer

Expert verified
The graph of \(y=(x-2)^2-1\) will be a parabola that opens upwards with vertex at \((2, -1)\) and axis of symmetry at \(x=2\)

Step by step solution

01

Identify the vertex of the quadratic function

The vertex is given as (h, k) in the general form of the function. Here h = 2 (which shifts the graph 2 places to the right) and k = -1 (which shifts the graph 1 place down). So the vertex of the quadratic is \((2,-1)\)
02

Identify the Axis of symmetry

The axis of symmetry of a parabola is the vertical line \(x = h\). In this case, the axis of symmetry is \(x = 2\)
03

Determine the direction of the parabola

Since the coefficient of \(x^2\) is positive in the provided function, the parabola will open upwards

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Key Concepts

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

Vertex Form
The vertex form of a quadratic function is a very useful expression that allows us to easily identify the vertex of the parabola it represents. It is given by the formula \[y = a(x-h)^2 + k\].In this formula:
  • \(a\) determines the width and the direction of the parabola.
  • \(h\) and \(k\) are the coordinates of the vertex.
The vertex is the highest or lowest point on the parabola, depending on whether it opens downwards or upwards. This form makes it simple to graph a quadratic function because knowing the vertex gives you a starting point for sketching the curve.
For example, in the function \(y=(x-2)^2-1\), the vertex is \((2, -1)\). This shows that the parabola shifts 2 units to the right and 1 unit down from the origin.
Parabola
A parabola is a U-shaped graph that is symmetrical. It is the graphical representation of a quadratic function. Parabolas have several unique features that distinguish their graph:
  • They have a single axis of symmetry.
  • They open either upwards or downwards.
  • They always have a vertex, which is either the highest or lowest point of the graph.
Parabolas can appear different based on their equation's coefficients, particularly the coefficient \(a\) in the standard form. When the coefficient is positive, as in our function, the parabola opens upwards like a cup. If it were negative, it would open downwards like an upside-down cup. A parabola will infinitely extend, but understanding these properties helps in sketching and working with these curves effectively.
Axis of Symmetry
The axis of symmetry of a parabola is a crucial concept because it divides the parabola into two mirror-image halves. For quadratic functions in vertex form, \(y = a(x-h)^2 + k\), the axis of symmetry is the vertical line given by \(x = h\).In simpler terms, the axis of symmetry is always a vertical line that passes through the vertex of the parabola. This line is critical in graphing the parabola because every point on one side of it mirrors a point on the other side.
For instance, in our example \(y=(x-2)^2-1\), the axis of symmetry is \(x=2\). This informs us that if the parabola is folded along this line, the two halves would coincide.
Graph Shifting
Graph shifting refers to moving the entire graph of a function in the coordinate plane. In the case of a quadratic function in vertex form, shifts can occur both horizontally and vertically.
Horizontal shifts are determined by the value of \(h\) in the vertex form equation \(y = a(x-h)^2 + k\):
  • If \(h\) is positive, the parabola shifts to the right.
  • If \(h\) is negative, the parabola shifts to the left.
Vertical shifts depend on the value of \(k\):
  • If \(k\) is positive, the graph moves up.
  • If \(k\) is negative, it moves down.
In our example \(y=(x-2)^2-1\), \(h = 2\) and \(k = -1\) indicate the graph moves 2 units to the right and 1 unit downward. This transforming of the graph's position while maintaining its shape is an essential skill in graphing functions.

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Most popular questions from this chapter

reflection in the \(y\)-axis followed by a translation 4 units right $$ \begin{aligned} h(x) &=f(-x) \\ &=2(-x)^2+6(-x) \\ &=2(x)^2-6 x \\ g(x) &=h(x-4) \\ &=2(x-4)^2+6(x-4) \\ &=2 x^2-10 x+8 \end{aligned} $$

\(\frac{2}{3}=\frac{x}{9}\)

\(f(x)=-4(x+1)^2-5\)

\(f(x)=\frac{3}{2} x^2+6 x+4\)

\(f(x)=0.4(x-1)^2\)
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