/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 Find the vertex, the \(x\) -inte... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the vertex, the \(x\) -intercepts (if any), and sketch the parabola. \(f(x)=-2 x^{2}+6 x-3\)

Short Answer

Expert verified
The vertex of the parabola is \((\frac{3}{2}, -\frac{3}{2})\), the \(x\)-intercepts are \(x = \frac{-6 + \sqrt{12}}{-4}\) and \(x = \frac{-6 - \sqrt{12}}{-4}\), and the graph is a downward-opening parabola with these vertex and x-intercepts.

Step by step solution

01

Find the vertex of the parabola

The parabola is given by the equation \(f(x) = -2x^2 + 6x - 3\). To find the vertex, we'll use the vertex formula as follows: \(x_{vertex} = \frac{-b}{2a}\) Here, \(a = -2, b = 6\), so \(x_{vertex} = \frac{-6}{2 \times (-2)} = \frac{6}{4} = \frac{3}{2}\) Now, plug the \(x_{vertex}\) value back into the equation to find the y-coordinate of the vertex: \(y_{vertex} = -2\left(\frac{3}{2}\right)^2 + 6\left(\frac{3}{2}\right) - 3\) \(y_{vertex} = -2\left(\frac{9}{4}\right) + 9 - 3\) \(y_{vertex} = -\frac{9}{2} + 6 = -\frac{3}{2}\) So, the vertex of the parabola is \((\frac{3}{2}, -\frac{3}{2})\).
02

Find the \(x\)-intercepts

To find the \(x\)-intercepts, we need to set the function equal to 0 and solve for \(x\). \(0 = -2x^2 + 6x - 3\) The quadratic is not factorable, so we will need to use the quadratic formula to find the roots: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) Plug the values \(a = -2, b = 6, c = -3,\) into the formula, and we obtain: \(x = \frac{-6 \pm \sqrt{6^2 - 4(-2)(-3)}}{2(-2)}\) \(x = \frac{-6 \pm \sqrt{36 - 24}}{-4}\) \(x = \frac{-6 \pm \sqrt{12}}{-4}\) We get two x-intercepts at \(x = \frac{-6 + \sqrt{12}}{-4}\) and \(x = \frac{-6 - \sqrt{12}}{-4}\).
03

Sketch the parabola

Now that we have the vertex and the \(x\)-intercepts, we can sketch the parabola. 1. Plot the vertex at \((\frac{3}{2}, -\frac{3}{2})\). 2. Plot the \(x\)-intercepts at \(x = \frac{-6 + \sqrt{12}}{-4}\) and \(x = \frac{-6 - \sqrt{12}}{-4}\). 3. Since the coefficient of the \(x^2\) term is negative, the parabola will open downwards. Now, connect the vertex and the x-intercepts with a smooth, downward-opening curve. In summary, the vertex of the parabola is \((\frac{3}{2}, -\frac{3}{2})\), the \(x\)-intercepts are \(x = \frac{-6 + \sqrt{12}}{-4}\) and \(x = \frac{-6 - \sqrt{12}}{-4}\), and the graph shows a downward-opening parabola with the given vertex and x-intercepts.

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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 the point where the graph changes direction. It can be the highest or lowest point on the graph, depending on the parabola's orientation. For a parabola in the form, \( f(x) = ax^2 + bx + c \), the vertex can be found using a specific formula for the x-coordinate of the vertex:
  • \( x_{vertex} = \frac{-b}{2a} \)
This formula uses the coefficients of the quadratic expression, where \( a \) is the coefficient of \( x^2 \), and \( b \) is the coefficient of \( x \). Once you calculate \( x_{vertex} \), plug it back into the equation to find the y-coordinate of the vertex:
  • \( y_{vertex} = f(x_{vertex}) \)
This gives the vertex as the point \((x_{vertex}, y_{vertex})\). For example, in the function \( f(x) = -2x^2 + 6x - 3 \), the vertex calculated is \( (\frac{3}{2}, -\frac{3}{2}) \). This vertex is crucial because it tells us both the direction of the graph and its extremum point.
Quadratic Formula
The quadratic formula is a powerful tool used to find the roots of a quadratic equation, which are the x-values at which the graph of the equation crosses the x-axis. A quadratic equation is generally expressed in the form \( ax^2 + bx + c = 0 \). The quadratic formula provides a direct solution for \( x \):
  • \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
The term under the square root sign, \( b^2 - 4ac \), is called the discriminant. It determines the nature of the roots:
  • If \( b^2 - 4ac > 0 \), there are two distinct real roots.
  • If \( b^2 - 4ac = 0 \), there is one real root (also called a repeated or double root).
  • If \( b^2 - 4ac < 0 \), there are no real roots but rather two complex roots.
For the function \( f(x) = -2x^2 + 6x - 3 \), using the quadratic formula gives the solutions for the x-intercepts: \( x = \frac{-6 \pm \sqrt{12}}{-4} \). This calculation is fundamental for determining how the parabola intersects the x-axis.
X-Intercepts
X-intercepts, also known as roots or zeros, are the points where a graph crosses the x-axis. For any function, finding the x-intercepts involves determining the x-values for which the function equals zero. For a quadratic function \( f(x) = ax^2 + bx + c \), you set \( f(x) = 0 \) and solve for \( x \).The x-intercepts can give vital information about the parabola's behavior and can be found using factoring (if possible), completing the square, or more commonly, the quadratic formula as described previously. For the quadratic \( f(x) = -2x^2 + 6x - 3 \), the x-intercepts are the solutions \( x = \frac{-6 + \sqrt{12}}{-4} \) and \( x = \frac{-6 - \sqrt{12}}{-4} \). These tell us where the parabola touches or crosses the x-axis.Knowing the position of x-intercepts within the graph helps visualize and sketch the shape of the parabola effectively, helping us understand the function's practical applications.

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