Problem 6
Question
The graph of a quadratic function is given. (a) Find the coordinates of the vertex. (b) Find the maximum or minimum value of f. (c) Find the domain and range of f. $$ f(x)=-x^{2}+6 x-5 $$
Step-by-Step Solution
Verified Answer
(a) Vertex is (3, 4). (b) The maximum value is 4. (c) Domain is \((-\infty, \infty)\) and range is \((-\infty, 4]\).
1Step 1: Identify the form of the quadratic function
The given quadratic function is presented in the standard form: \[ f(x) = ax^2 + bx + c \] where \( a = -1 \), \( b = 6 \), and \( c = -5 \). This form will help us find the vertex coordinates and determine if the function has a maximum or minimum value.
2Step 2: Calculate the vertex's x-coordinate
The formula for the x-coordinate of the vertex of a quadratic function \( ax^2 + bx + c \) is given by: \[ x = -\frac{b}{2a} \] Substitute \( b = 6 \) and \( a = -1 \): \[ x = -\frac{6}{2(-1)} = 3 \] So, the x-coordinate of the vertex is 3.
3Step 3: Calculate the vertex's y-coordinate
To find the y-coordinate of the vertex, substitute \( x = 3 \) back into the function: \[ f(3) = -(3)^2 + 6(3) - 5 \] \[ = -9 + 18 - 5 \] \[ = 4 \] Thus, the y-coordinate of the vertex is 4. The vertex of the parabola is at the point (3, 4).
4Step 4: Determine the maximum or minimum value
Since the parabola opens downwards (as indicated by \( a = -1 < 0 \)), the vertex represents the maximum point. Therefore, the maximum value of \( f(x) \) is the y-coordinate of the vertex, which is 4.
5Step 5: Find the Domain of the function
A quadratic function is defined for all real numbers, so the domain of \( f(x) = -x^2 + 6x - 5 \) is all real numbers, denoted as \( (-\infty, \infty) \).
6Step 6: Find the Range of the function
Since this is a downward-opening parabola with a vertex at (3, 4), the range of \( f(x) \) is all y-values less than or equal to the y-coordinate of the vertex.Thus, the range is \( (-\infty, 4] \).
Key Concepts
Vertex of a Quadratic FunctionMaximum or Minimum ValueDomain and Range of Functions
Vertex of a Quadratic Function
The vertex of a quadratic function serves as a central point on its graph. This point can be a maximum or a minimum, depending on the direction in which the parabola opens. For the standard form of a quadratic function, \[ f(x) = ax^2 + bx + c \], the vertex's x-coordinate can be calculated using this formula:
\[ x = -\frac{6}{2(-1)} = 3 \]Thus, the x-coordinate of the vertex is 3. To find the corresponding y-coordinate, substitute \( x = 3 \) back into the original equation:
- \(-\frac{b}{2a}\)
\[ x = -\frac{6}{2(-1)} = 3 \]Thus, the x-coordinate of the vertex is 3. To find the corresponding y-coordinate, substitute \( x = 3 \) back into the original equation:
- \( f(3) = -(3)^2 + 6(3) - 5 \)
- \( f(3) = -9 + 18 - 5 = 4 \)
Maximum or Minimum Value
The maximum or minimum value of a quadratic function corresponds directly to the vertex of the parabola. If the parabola opens upwards (where \( a > 0 \)), the vertex is the minimum point. Conversely, if the parabola opens downwards (where \( a < 0 \)), the vertex serves as the maximum point. This is because the direction in which the parabola opens determines the shape of the curve.
In our example, the function \( f(x) = -x^2 + 6x - 5 \) has \( a = -1 \). Since \( a < 0 \), the parabola opens downwards. This means the vertex at (3, 4) is a maximum point. Thus, the function reaches its highest value at the vertex, and this maximum value is 4. Knowing whether a point is a maximum or minimum helps in understanding the behavior of the function and its graph.
In our example, the function \( f(x) = -x^2 + 6x - 5 \) has \( a = -1 \). Since \( a < 0 \), the parabola opens downwards. This means the vertex at (3, 4) is a maximum point. Thus, the function reaches its highest value at the vertex, and this maximum value is 4. Knowing whether a point is a maximum or minimum helps in understanding the behavior of the function and its graph.
Domain and Range of Functions
Understanding the domain and range of functions is crucial in grasping the full picture of their behavior. The domain of a quadratic function includes all possible x-values, reflecting all inputs the function can accept. For a standard quadratic function, the domain is always all real numbers, or \((-\infty, \infty)\).
The range, however, reflects the output values, or y-values, that the function can produce. It is determined largely by the vertex and the parabola's direction. A downward-opening parabola, like our function \( f(x) = -x^2 + 6x - 5 \), peaks at the vertex. Thus, the range includes all y-values less than or equal to the y-coordinate of the vertex.
The range, however, reflects the output values, or y-values, that the function can produce. It is determined largely by the vertex and the parabola's direction. A downward-opening parabola, like our function \( f(x) = -x^2 + 6x - 5 \), peaks at the vertex. Thus, the range includes all y-values less than or equal to the y-coordinate of the vertex.
- For \( f(x) = -x^2 + 6x - 5 \), the range is \((-\infty, 4]\).
Other exercises in this chapter
Problem 6
\(3-8=\) Two polynomials \(P\) and \(D\) are given. Use either synthetic or long division to divide \(P(x)\) by \(D(x),\) and express \(P\) in the form \(P(x)=D
View solution Problem 6
A polynomial P is given. (a) Find all zeros of P, real and complex. (b) Factor P completely. \(P(x)=x^{5}+9 x^{3}\)
View solution Problem 7
List all possible rational zeros given by the Rational Zeros Theorem (but don' t check to see which actually are zeros). $$ R(x)=2 x^{5}+3 x^{3}+4 x^{2}-8 $$
View solution Problem 7
\(3-8=\) Two polynomials \(P\) and \(D\) are given. Use either synthetic or long division to divide \(P(x)\) by \(D(x),\) and express \(P\) in the form \(P(x)=D
View solution