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Chapter 2: Problem 26

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). $$ f(x)=x^{2}-30 x+225 $$

Short Answer

Expert verified
The quadratic function in standard form is \(f(x)=(x-15)^2\). The graph is a parabola with vertex at (15, 0), an axis of symmetry at \(x = 15\), and an x-intercept at \(x = 15\).

Step by step solution

01

Converting to Standard Form

First, let's convert the given function to standard form. This is done by completing the square. Given the function: \(f(x)=x^{2}-30 x+225\), the constant term inside the square should be \(b/2)^2 = (30/2)^2 =225\). So the function in standard form is \(f(x)=(x-15)^2\)
02

Identifying the Vertex

The vertex of the parabola in standard form is found at the point \( (h, k) \). Since our function is \(f(x)=(x-15)^2\), our h and k values are \(15\) and \(0\) respectively. Therefore, the vertex is at point (15, 0).
03

Identifying the Axis of Symmetry

The axis of symmetry for a function in standard form is the vertical line \(x = h\). For our function, that means the axis of symmetry is the vertical line \(x = 15\).
04

Finding the x-intercepts

The x-intercepts are found by setting \(f(x)\) to \(0\) and solving for \(x\). Therefore, 0 = (x-15)^2. By solving, the x-intercept is \(x = 15\).

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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 Function
Understanding the standard form of a quadratic function is crucial for graphing parabolas. The standard form is expressed as \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( a \) is not equal to zero. When dealing with quadratic functions, a key step is to express them in this format. This reveals several characteristics such as the direction of the opening of the parabola (upward if \( a > 0\), downward if \( a < 0\)) and the y-intercept at \( c \).

For the exercise, \( f(x) = x^2 - 30x + 225 \), the function is already in standard form but further simplification through completing the square, as shown in the step-by-step solution, gives us a more precise view of the parabola's graph.
Vertex of a Parabola
The vertex of a parabola represents the highest or lowest point of the curve, depending on its orientation. In standard form, \( f(x) = (x-h)^2 + k \), the vertex to be found at the point \( (h, k) \).

In our example, after converting the function into vertex form by completing the square, we get \( f(x) = (x - 15)^2 \). It is then easy to spot that \( h = 15 \) and \( k = 0 \), making the vertex at the point \( (15, 0) \). This point is crucial for sketching the parabola, as it acts as a reference from which the shape of the curve is drawn.
Axis of Symmetry
The axis of symmetry of a parabola is a vertical line that divides the curve into two mirror images. For a parabola in the standard form \( f(x) = (x-h)^2 + k \), the axis of symmetry is given by the equation \( x = h \).

Referring back to our exercise, with \( f(x) = (x - 15)^2 \), the axis of symmetry is \( x = 15 \). This line passes through the vertex and is a pivotal guide in ensuring the two halves of the parabola are symmetrical when graphing.
X-intercepts of a Parabola
X-intercepts of a parabola, also known as the roots or zeros, are the points where the parabola crosses the x-axis. To find these points, we set the quadratic equation equal to zero and solve for \( x \).

From the step-by-step solution, we know that the given function \( f(x) = (x-15)^2 \), when set to \( 0 \), gives us \( x = 15 \) after solving. This means there is only one x-intercept at \( x = 15 \), which also happens to coincide with the vertex. In cases where the quadratic formula or factoring is needed, there may be two distinct x-intercepts, or none if the parabola does not cross the x-axis.

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