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Chapter 2: Problem 32

Using Standard Form to Graph a Parabola In Exercises \(17-34\) , write the quadratic function in standard form and sketch its graph. Identify the vertex, axis of symmetry, and \(x\) -intercept(s). $$ f(x)=2 x^{2}-x+1 $$

Short Answer

Expert verified
The quadratic function in standard form is \(f(x)=(x-\frac{1}{4})^{2}+\frac{7}{16}\). The vertex is \(\left(\frac{1}{4}, \frac{7}{16}\right)\), the axis of symmetry is \(x=\frac{1}{4}\), and there are no x-intercepts.

Step by step solution

01

Write the Equation in Standard Form (Complete the Square)

The standard form of a quadratic equation is given by \( f(x) = a(x - h)^2 + k \) where (h, k) is the vertex of the parabola. Let's write the given function in the standard form. Starting with \( f(x)=2x^{2}-x+1 \), the standard form can be obtained by completing the square. Divide every term of f(x) by 2: \( f(x)=x^{2}-\frac{1}{2}x+\frac{1}{2} \). Now complete the square by adding and subtracting \(\left(\frac{1}{2} * \frac{1}{2}\right)^{2}\) inside the parenthesis: \( f(x)=\left(x^{2}-\frac{1}{2}x+ \left(\frac{1}{2} * \frac{1}{2}\right)^{2} \right) - \left(\frac{1}{2} * \frac{1}{2}\right)^{2} + \frac{1}{2} \). Simplifying gives \( f(x)=\left(x-\frac{1}{4}\right)^{2}+\frac{7}{16} \)
02

Determine the Vertex, Axis of Symmetry, and X-Intercepts

The vertex of the parabola is given by the (h, k) in the standard form of the function. Thus, the vertex is \(\left(\frac{1}{4}, \frac{7}{16}\right)\). The axis of symmetry is a vertical line through the vertex, so it has the equation \(x=h\), which means the axis of symmetry is \(x=\frac{1}{4}\). To find the x-intercepts, set \(f(x) = 0\) and solve for x. However, since our f(x) is always positive (due to the squared term), there are no x-intercepts.
03

Plot the Graph

Using the information from Step 2, start by plotting the vertex and the axis of symmetry. Then since this is a standard upward-opening parabola (coefficient of x^2 is positive), plot two symmetrical points on either side of the vertex to get the general shape. With vertex at \(\left(\frac{1}{4}, \frac{7}{16}\right)\) and no x-intercepts, the graph would be a U-shaped curve opening upward with vertex slightly to the right from y-axis, just above the x-axis. And, it would not cross the x-axis due to no x-intercepts.

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Key Concepts

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

Standard Form of a Quadratic Equation
Understanding the standard form of a quadratic equation is a foundational skill in algebra. A quadratic equation is typically represented by the equation \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( a \) is not zero. This form allows us to quickly identify the coefficients that influence the shape and position of a parabola on a graph.
  • The coefficient \( a \) determines the opening direction and the width of the parabola; if \( a \) is positive, the parabola opens upwards, and if negative, it opens downwards.
  • The coefficients \( b \) and \( c \) affect the location of the parabola along the x-axis and y-axis, respectively.
When graphing, one of the first steps is to format the quadratic equation into this standard structure to easily extract critical information for plotting the parabola.
Completing the Square
Completing the square is a powerful technique used to rewrite a quadratic equation in a form that makes it easier to identify the vertex.
To complete the square for an equation in the form \( ax^2 + bx + c \), one would:
  • Ensure the coefficient of \( x^2 \) is 1. If not, divide the entire equation by \( a \).
  • Rearrange the equation to isolate the \( x \) terms: \( x^2 + (b/a)x = -c/a \).
  • Add and subtract \( (b/2a)^2 \) on the left-hand side to complete the square.
  • Rewrite the equation as \( (x + b/2a)^2 = (b/2a)^2 - c/a \).
  • Finally, express the equation in the vertex form: \( f(x) = a(x - h)^2 + k \), where \( h = -b/2a \) and \( k \) is the value on the right-hand side.
By completing the square, we transform the quadratic equation into a form where the vertex and axis of symmetry become apparent, which is invaluable for graphing.
Vertex of a Parabola
The vertex of a parabola is a crucial point where the curve changes direction. In the context of the standard quadratic equation \( f(x) = a(x - h)^2 + k \), after completing the square, the vertex \( (h, k) \) becomes easily identifiable.
  • The \( h \) value indicates where the parabola crosses the axis of symmetry on the x-axis.
  • The \( k \) value represents the minimum or maximum point of the parabola on the y-axis, depending on whether it opens upwards or downwards.
The formula to find the vertex from the standard form \( ax^2 + bx + c \) without completing the square involves calculating \( h = -b/2a \) and substituting \( h \) back into the original equation to find \( k \). When plotting a graph, the vertex is the first point you'll want to plot as it gives you a reference for the shape and position of the parabola.
Axis of Symmetry
The axis of symmetry in a parabola is a vertical line that divides the curve into two mirror-image halves. For the quadratic function \( f(x) = a(x - h)^2 + k \), the axis of symmetry is the line \( x = h \).
  • For every point on the parabola, there is another point reflected across the axis of symmetry that has the same y-value.
  • Knowing the axis of symmetry is particularly useful because it helps in graphing the parabola accurately.
  • It also serves as a guide for determining the x-coordinates of additional points on the parabola when plotting the graph.
When graphing a quadratic function, after locating the vertex, draw the axis of symmetry to serve as a helpful reference for the rest of the graph.
X-Intercepts
X-intercepts are the points where the parabola crosses the x-axis. They are also known as 'roots' or 'zeroes' of the quadratic function. To find the x-intercepts algebraically, one must set the function equal to zero, \( f(x) = 0 \), and solve for x.
  • If the quadratic equation can be factored, the intercepts can be found by setting each factor equal to zero.
  • If factoring is not possible, one can use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), to find the intercepts.
  • In some cases, the parabola does not cross the x-axis and therefore has no real x-intercepts, which indicates that the function has no real roots.
Graphically, the x-intercepts are vital points for sketching the curve of the parabola. They provide information about the spread and orientation of the parabola on the graph.

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