Problem 35
Question
Describe the graph of the quadratic function. Identify the vertex and \(x\) -intercept(s). Use a graphing utility to verify your results. \(f(x)=-2 x^{2}+16 x-31\)
Step-by-Step Solution
Verified Answer
The vertex of the function is at the point \((4, -1)\) and the x-intercepts are at points \((-3, 0)\) and \((31/2, 0)\). This is confirmed by the graph of the function.
1Step 1: Finding the Vertex
In a quadratic function in standard form \(f(x) = ax^2 + bx + c\), the vertex is given by the point \((-b/2a, f(-b/2a))\). Here, \(a = -2\) and \(b = 16\), so substituting these values, we find that the x-coordinate of the vertex is \(-b/2a = -16/(-4) = 4\). Substituting \(x = 4\) into the equation, we get the y-coordinate, \(f(4) = -2*4^2 + 16*4 - 31 = -1\). So, the vertex is at the point \((4, -1)\).
2Step 2: Finding the x-intercepts
The x-intercepts of the function are given by the solutions to the equation \(f(x) = 0\), i.e., \(-2x^2 + 16x - 31 = 0\). This equation can be solved using the quadratic formula \(x = [-b±sqrt(b^2 - 4ac)]/(2a)\). Substituting \(a = -2\), \(b = 16\), and \(c = -31\) into the formula, we get the two solutions \(x = -3\) and \(x = 31/2\). So, the x-intercepts are \((-3, 0)\) and \((31/2, 0)\).
3Step 3: Verifying with a Graphing Utility
Through the use of a graphing utility, one can plot the function \(f(x) = -2x^2 + 16x - 31\). Looking at the graph, it can be confirmed that the vertex is indeed at \((4, -1)\), and that the function intersects the x-axis at \((-3, 0)\) and \((31/2, 0)\).
Key Concepts
Vertex of a QuadraticX-Intercepts of a QuadraticQuadratic FormulaGraphing Utility Verification
Vertex of a Quadratic
The vertex of a quadratic function is a pivotal concept in graphing parabolas. It represents the highest or lowest point on the graph, depending on whether the parabola opens upwards or downwards. In the case of the function
Remember, the vertex is key to understanding the shape and position of the parabola. It can also help us determine the function's maximum or minimum value, which in this case is a minimum at
f(x)=-2x^{2}+16x-31, we can determine the vertex by using the formula (-b/2a, f(-b/2a)). With a=-2 and b=16, we calculate the x-coordinate of the vertex as -b/2a = -16/(-4) = 4. Subsequently, we find the y-coordinate by substituting x = 4 into the function, yielding f(4) = -1. Thus, the vertex is located at (4, -1).Remember, the vertex is key to understanding the shape and position of the parabola. It can also help us determine the function's maximum or minimum value, which in this case is a minimum at
y = -1. Graphically, if you were sketching the parabola, you would start at the vertex and work your way symmetrically on both sides to plot the curve.X-Intercepts of a Quadratic
Finding the x-intercepts of a quadratic function is a common algebraic exercise that reveals where the parabola crosses the x-axis. These intercepts are the solutions to the equation
These solutions give us the x-intercepts as points
f(x)=0. For the quadratic function f(x)=-2x^{2}+16x-31, we can solve for the x-intercepts using the quadratic formula x = [-b±sqrt(b^2 - 4ac)]/(2a). With our given coefficients, a=-2, b=16, and c=-31, we find two solutions: x = -3 and x = 31/2.These solutions give us the x-intercepts as points
(-3, 0) and (31/2, 0). Remember, x-intercepts are also referred to as zeros or roots of the function, and they provide a clear visual when looking at the function's graph.Quadratic Formula
The quadratic formula, an essential tool in algebra, allows us to find the solutions to any quadratic equation of the form
In our function
ax^{2} + bx + c = 0. The formula is written as x = [-b±sqrt(b^2 - 4ac)]/(2a). It provides the roots by accounting for both the possibility of two real distinct solutions, one real double solution, or two complex solutions depending on the discriminant (b^2 - 4ac).In our function
f(x)=-2x^{2}+16x-31, by substituting the coefficients into the quadratic formula, we derived the x-intercepts. The ability to compute these solutions algebraically is crucial, especially when graphing is impractical or when an exact numerical answer is required. The quadratic formula is a reliable method that applies to all forms of quadratic equations.Graphing Utility Verification
After calculating the vertex and x-intercepts of a quadratic function algebraically, it's beneficial to use a graphing utility for verification. A graphing utility allows you to visualize the curve of the function and confirm the accuracy of your solutions. When we input the quadratic function
Utilizing graphing technology is a powerful way to double-check your work and to gain a deeper understanding of the function's graphical behavior. It can also reveal additional features of the quadratic function, such as the direction of opening, the axis of symmetry, and the general steepness or width of the parabola.
f(x)=-2x^{2}+16x-31 into a graphing utility, it will display the parabola, illustrating that it intersects the x-axis at the points we calculated: (-3, 0) and (31/2, 0), and that the vertex appears at the point (4, -1).Utilizing graphing technology is a powerful way to double-check your work and to gain a deeper understanding of the function's graphical behavior. It can also reveal additional features of the quadratic function, such as the direction of opening, the axis of symmetry, and the general steepness or width of the parabola.
Other exercises in this chapter
Problem 35
Use a graphing utility to graph the two equations in the same viewing window. Use the graphs to verify that the expressions are equivalent. Verify the results a
View solution Problem 35
Find all the zeros of the function and write the polynomial as a product of linear factors. Use a graphing utility to verify your results graphically. (If possi
View solution Problem 35
Use the Leading Coefficient Test to describe the right-hand and left-hand behavior of the graph of the polynomial function. Use a graphing utility to verify you
View solution Problem 36
Determine the value that the function \(f\) approaches as the magnitude of \(x\) increases. Is \(f(x)\) greater than or less than this value when \(x\) is posit
View solution