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Chapter 2: Problem 8

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 \((5, 25)\), and intercepts are \((0, 0)\) and \((10, 0)\). The graph is a downward-opening parabola.

Step by step solution

01

Identify the Quadratic Function

The quadratic function is given as \( f(x) = -x^2 + 10x \). This function is in the form \( ax^2 + bx + c \) where \( a = -1 \), \( b = 10 \), and \( c = 0 \).
02

Express the Function in Standard Form

The standard form of a quadratic function is \( ax^2 + bx + c \). Since the given function \( f(x) = -x^2 + 10x \) already matches this form, it is already expressed in standard form.
03

Find the Vertex of the Quadratic Function

The vertex form of a quadratic function is \( y = a(x-h)^2 + k \), where \( (h, k) \) is the vertex. The vertex \( h \) can be found using the formula \( h = \frac{-b}{2a} \). Substituting \( a = -1 \) and \( b = 10 \), we calculate \( h = \frac{-10}{2(-1)} = 5 \). The \( k \)-value is found by substituting \( x = 5 \) back into the function: \( f(5) = -(5)^2 + 10(5) = 25 \). Thus, the vertex is \((5, 25)\).
04

Find the x-intercepts

The \( x \)-intercepts are found by setting \( f(x) = 0 \), i.e., \( -x^2 + 10x = 0 \). Factoring gives \( -x(x - 10) = 0 \), which yields \( x = 0 \) and \( x = 10 \). Thus, the \( x \)-intercepts are \((0, 0)\) and \((10, 0)\).
05

Find the y-intercept

The \( y \)-intercept is the value of \( f(x) \) when \( x = 0 \). From the function \( f(x) = -x^2 + 10x \), substitute \( x = 0 \) to find \( f(0) = -(0)^2 + 10(0) = 0 \). Thus, the \( y \)-intercept is \((0, 0)\).
06

Sketch the Graph of the Function

The graph of the quadratic function \( f(x) = -x^2 + 10x \) is a downward-opening parabola due to the negative coefficient of \( x^2 \). Mark the vertex at \((5, 25)\), and include the \( x \)-intercepts \((0, 0)\) and \((10, 0)\) as well as the \( y \)-intercept \((0, 0)\) on the graph. The parabola should open downwards.

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Key Concepts

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

Vertex of a Parabola
The vertex of a parabola is a crucial point that gives insight into the parabola's highest or lowest point, depending on its orientation. For a quadratic function like \[ f(x) = ax^2 + bx + c \]the vertex can be found using the formula: \[ h = \frac{-b}{2a} \]where "h" is the x-coordinate of the vertex. Once "h" is found, substitute it back into the function to find the y-coordinate "k", completing the vertex as the point \[ (h, k) \].
The vertex indicates:
  • The maximum or minimum value of the function.
  • The peak of the graph if the parabola opens downwards (like \( f(x) = -x^2 + 10x \)).
  • The direction of the parabola, determined by the sign of \( a \).
In our example, we calculated \( h = 5 \) and the vertex is \( (5, 25) \). This tells us the highest point on the parabola since it opens downward.
x-intercepts
The x-intercepts of a quadratic function are the points where the graph crosses the x-axis. To find them, set the function equal to zero and solve for \( x \):\[-x^2 + 10x = 0\].
Factoring helps here, so \[-x(x - 10) = 0\]which gives x-intercepts at \( x = 0 \) and \( x = 10 \). These translate to points \( (0, 0) \) and \( (10, 0) \) on the graph.
x-intercepts are:
  • Crucial for determining where the function equals zero.
  • Points of intersection with the x-axis, helping guide your graph.
Remember, a parabola can have two, one, or no real x-intercepts depending on its position and orientation.
y-intercepts
The y-intercept is where the graph of the function crosses the y-axis. This occurs when \( x = 0 \). To find the y-intercept, substitute \( x = 0 \) into the quadratic function.
For example, with the function \[ f(x) = -x^2 + 10x \], substitute to get \[ f(0) = -(0)^2 + 10(0) = 0 \].
Hence, the y-intercept here is \( (0, 0) \).
The y-intercept is:
  • Always a single point unless the function is degenerate (not a true parabola).
  • Useful in providing a starting view for the graph.
  • A fixed point all parabolas will pass through when graphed.
Graphing Parabolas
Graphing a quadratic function means illustrating its set of points, which form a parabola. Here's a step-by-step method for the function \[ f(x) = -x^2 + 10x \]:
  • Identify if the parabola opens upwards or downwards. Here, it's downwards, as the coefficient of \( x^2 \) is negative.
  • Plot the vertex, which is \( (5, 25) \).
  • Mark the x-intercepts at \( (0, 0) \) and \( (10, 0) \).
  • Add the y-intercept, \( (0, 0) \), although it's already plotted as an x-intercept.
Connect these points smoothly in a U-shaped curve (inverted for downward-opening).
Ensure the graph is symmetric about the vertex, with equal spacing between intercepts confirming accuracy.

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