Problem 51 Graph each function using the ve... [FREE SOLUTION]

Chapter 12: Problem 51

Graph each function using the vertex formula. Include the intercepts. \(f(x)=\frac{1}{2} x^{2}-4 x+5\)

Short Answer

Expert verified

The vertex of the function \(f(x) = \frac{1}{2}x^2 - 4x + 5\) is at \((4, 8)\). The x-intercepts are at \(x = 4 - \sqrt{6}\) and \(x = 4 + \sqrt{6}\), and the y-intercept is at (0, 5).

Step by step solution

Determine a, b, and c from the quadratic function

Given the function \(f(x) = \frac{1}{2}x^2 - 4x + 5\), we can identify the coefficients from the standard form of a quadratic function, which is: \(f(x) = ax^2 + bx + c\) For the given function, \(a = \frac{1}{2}\), \(b = -4\), and \(c = 5\).

Find the vertex

To find the vertex of the quadratic function, we will use the formula for the vertex: \(v(x) = \left(h, k\right) = \left(\frac{-b}{2a}, f\left(\frac{-b}{2a}\right)\right)\) Plugging in the values for a and b, we find: \(h = \frac{-(-4)}{2(\frac{1}{2})} = \frac{4}{1} = 4\) Now, we need to find the value of the function at h: \(k = f(h) = f(4)\). Plugging in x = 4 into the function, we get: \(k = f(4) = \frac{1}{2}(4)^2 - 4(4) + 5 = 2(16) - 16 + 5 = 8\) So, the vertex of the function is at \((4, 8)\).

Find the x-intercepts

To find the x-intercepts, we need to set the function equal to 0: \(\frac{1}{2}x^2 - 4x + 5 = 0\) Use the quadratic formula \(\frac{-b \pm \sqrt{b^2-4ac}}{2a}\), and plug in the values for a, b, and c: \(x = \frac{4 \pm \sqrt{(-4)^2 - 4(\frac{1}{2})(5)}}{2(\frac{1}{2})} \) \(x = \frac{4 \pm \sqrt{16 - 10}}{1}\) \(x = \frac{4 \pm \sqrt{6}}{1}\) So, the x-intercepts are at \(x = 4 - \sqrt{6}\) and \(x = 4 + \sqrt{6}\).

Find the y-intercept

To find the y-intercept of the function, we need to plug in x = 0: \(f(0) = \frac{1}{2}(0)^2 - 4(0) + 5 = 5\) So, the y-intercept is at (0, 5). Now that we have the vertex, x-intercepts, and y-intercept, we can sketch the graph of the function \(f(x) = \frac{1}{2}x^2 - 4x + 5\).

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Key Concepts

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

Vertex Formula

The vertex of a quadratic function is a significant point that helps in understanding the function's graph. The vertex formula is part of the vertex form of a quadratic equation:

A quadratic function can be expressed in the form \(f(x) = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants.
The vertex \((h, k)\) is found using \(h = \frac{-b}{2a}\) and \(k = f(h)\).

To find the vertex of a function like \(f(x) = \frac{1}{2}x^2 - 4x + 5\), we substitute the coefficients into the formula:

For \(a = \frac{1}{2}\) and \(b = -4\), we calculate \(h = \frac{-(-4)}{2 \times \frac{1}{2}} = 4\).
To find \(k\), evaluate the function at \(h\): \(k = f(4) = \frac{1}{2}(4)^2 - 4(4) + 5 = 8\).
The vertex is \((4, 8)\).

Graphing Quadratic Functions

Graphing a quadratic function like \(f(x) = \frac{1}{2}x^2 - 4x + 5\) involves several steps and key points that help visualize the curve:

The vertex, previously found using the vertex formula, is the pinnacle of the parabola for standard form quadratics.
The graph is a parabola that opens upwards if \(a > 0\) (as with our function where \(a = \frac{1}{2}\)), or downwards if \(a < 0\).
The vertex \((4, 8)\) determines the axis of symmetry, \(x=4\).

After locating the vertex and determining the direction of the parabola, plot additional points using the intercepts and symmetric points around the vertex. This helps you sketch the curve accurately.

X-Intercepts

The x-intercepts of a quadratic function are the points where the graph crosses the x-axis, meaning the output value \(f(x)\) is zero.

To find the x-intercepts, set the function equal to zero: \(\frac{1}{2}x^2 - 4x + 5 = 0\).
Use the quadratic formula to solve for \(x\): \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = \frac{1}{2}\), \(b = -4\), and \(c = 5\).
Calculating gives us: \(x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(\frac{1}{2})(5)}}{2 \times \frac{1}{2}}\).
Simplifying yields: \(x = 4 \pm \sqrt{6}\).
The x-intercepts are approximately at \(x = 4 + \sqrt{6}\) and \(x = 4 - \sqrt{6}\).

Y-Intercepts

The y-intercept of a quadratic function is the point where the graph crosses the y-axis. This occurs when \(x = 0\).

To find the y-intercept, substitute \(x = 0\) into the quadratic function \(f(x) = \frac{1}{2}x^2 - 4x + 5\).
Evaluate: \(f(0) = \frac{1}{2}(0)^2 - 4(0) + 5 = 5\).
Thus, the y-intercept is the point \((0, 5)\) on the graph.

This point gives a simple starting place to begin graphing or checking work for accuracy. It shows where the function's parabola meets the y-axis.

91影视

Graph each function using the vertex formula. Include the intercepts. \(f(x)=\frac{1}{2} x^{2}-4 x+5\)

Short Answer

Step by step solution

Determine a, b, and c from the quadratic function

Find the vertex

Find the x-intercepts

Find the y-intercept

Key Concepts

Vertex Formula

Graphing Quadratic Functions

X-Intercepts

Y-Intercepts

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