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Chapter 4: Problem 12

A quadratic function is given. (a) Express the quadratic function in standard form. (b) Find its vertex and its x- and y-intercept(s). (c) Sketch its graph. $$ f(x)=-x^{2}+10 x $$

Short Answer

Expert verified
The vertex is (5, 25), x-intercepts are (0, 0) and (10, 0), and the y-intercept is (0, 0).

Step by step solution

01

Identify the Forms of a Quadratic Function

A quadratic function can be expressed in different forms. The standard form is given by \( ax^2 + bx + c \). Our function is given as \( f(x) = -x^2 + 10x \), which is already in standard form with \( a = -1 \), \( b = 10 \), and \( c = 0 \).
02

Find the Vertex

The vertex of a quadratic function in standard form \( ax^2 + bx + c \) can be found using the formula \(x = -\frac{b}{2a}\). Here, \( a = -1 \) and \( b = 10 \), so \( x = -\frac{10}{2 \times -1} = 5 \). Substitute back into the function to find the y-value: \( f(5) = -(5)^2 + 10 \times 5 = -25 + 50 = 25 \). Thus, the vertex is at \((5, 25)\).
03

Find the x-Intercepts

To find the x-intercepts, set \( f(x) = 0 \) and solve \(-x^2 + 10x = 0\). Factor out \(x\): \(x(-x + 10) = 0\). This gives solutions \( x = 0 \) and \( x = 10 \). Thus, the x-intercepts are \((0, 0)\) and \((10, 0)\).
04

Find the y-Intercept

The y-intercept of a function is found by evaluating \( f(0) \). Substitute \( x = 0 \) into the equation: \( f(0) = -(0)^2 + 10 \times 0 = 0 \). Thus, the y-intercept is \((0, 0)\).
05

Sketch the Graph

Plot the vertex \((5, 25)\), the x-intercepts \((0, 0)\) and \((10, 0)\), and the y-intercept \((0, 0)\). The parabola opens downwards because the coefficient of \(x^2\) is negative. Connect the points with a smooth curve to complete the graph.

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Key Concepts

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

Standard Form
The standard form of a quadratic function is an important structure that shows how a quadratic equation is organized. It is typically written as \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. Understanding this format is crucial because:
  • \( a \) determines the direction and width of the parabola. If \( a \) is positive, the parabola opens upwards, otherwise, it opens downwards.
  • \( b \) and \( c \) help locate the vertex and the axis of symmetry of the parabola.
If we consider the quadratic given, \( f(x) = -x^2 + 10x \), we can see it is in standard form with \( a = -1 \), \( b = 10 \), and \( c = 0 \). Notice how \( c \) being zero means there is no constant term involved, simplifying our calculation in finding intercepts and sketching graphs.
Vertex of a Parabola
Finding the vertex of a parabola is an essential step in understanding its properties. The vertex provides the highest or lowest point of the parabola, depending on its direction.The formula to find the x-coordinate of the vertex from the standard form \( ax^2 + bx + c \) is:\[ x = -\frac{b}{2a} \]With this formula, you can calculate the vertex's location. In the example equation, \( a = -1 \) and \( b = 10 \). Plugging these values in shows:\[ x = -\frac{10}{2 \cdot -1} = 5 \]Once you have \( x = 5 \), substitute it back to find the y-coordinate:\[ f(5) = -(5)^2 + 10 \cdot 5 = -25 + 50 = 25 \]Thus, the vertex is \((5, 25)\), indicating the peak of the downward-opening parabola.
X-intercepts
X-intercepts are the points where the parabola crosses the x-axis. They can be found by setting the quadratic equation equal to zero and solving for \( x \).For \( f(x) = -x^2 + 10x \):1. Solve \( -x^2 + 10x = 0 \).2. Factor out \( x \) as it is common in both terms: \[ x(-x + 10) = 0 \]3. This gives \( x = 0 \) or \( x = 10 \).These solutions correspond to the x-intercepts \((0, 0)\) and \((10, 0)\). Both points are vital in sketching the parabola, as they mark where the curve crosses the x-axis.
Y-intercepts
The y-intercept is where the graph of the quadratic function crosses the y-axis. You can find this by setting \( x = 0 \) in the function and calculating the result.For \( f(x) = -x^2 + 10x \), substitute \( x = 0 \):\[ f(0) = -(0)^2 + 10 \cdot 0 = 0 \]Thus, the y-intercept is clearly at \((0, 0)\). This point also coincides with one of the x-intercepts, making it a crucial point for graphing as it assures us where the parabola meets the origin.
Graphing Quadratic Functions
Graphing a quadratic function starts by plotting key features like the vertex and intercepts and understanding the parabola's general shape. Here are the steps to graph \( f(x) = -x^2 + 10x \):
  • Step 1: Plot the vertex \((5, 25)\).
  • Step 2: Mark the x-intercepts \((0, 0)\) and \((10, 0)\).
  • Step 3: Note the y-intercept, which is also \((0, 0)\), overlapping an x-intercept here.
  • Step 4: Since \( a = -1 \) is negative, draw the parabola opening downwards.
Connect these points smoothly to complete the graph. This visual representation helps in verifying and understanding the function's behavior, including the width and direction of the parabola.

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Most popular questions from this chapter

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