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Graphical Analysis In Exercises \(35-42,\) use a graphing utility to graph the quadratic function. Identify the vertex, axis of symmetry, and \(x\) -intercept(s). Then check your results algebraically by writing the quadratic function in standard form. $$ g(x)=\frac{1}{2}\left(x^{2}+4 x-2\right) $$

Short Answer

Expert verified
The vertex of the function is at (-2, -3), the axis of symmetry is at x= -2, and the x-intercepts are \((-2+sqrt(6), 0)\) and \((-2-sqrt(6), 0)\). The algebraic analysis agrees with these values.

Step by step solution

01

Compute the Vertex Square Form and Find the Vertex

The vertex form of a quadratic function is given by \( f(x)=a(x-h)^2+k \), where (h, k) are the coordinates of the vertex. Using the given function \( g(x)=0.5(x^2+4x-2) \), the term that completes the square within the parentheses is \( (b/2)^2 = (4/2)^2 = 4 \). Therefore, the function, when rewritten in vertex form becomes \( g(x)=0.5[(x+2)^2-6] \). Hence, the vertex, h is -2 and k is -3.
02

Identify the Axis of Symmetry

The quadratic function's axis of symmetry is \( x = h \). Given that h from the previous step is -2, the axis of symmetry is at \( x = -2 \).
03

Find the x-intercepts

The x-intercepts of the function are found by setting \( g(x) = 0 \). This gives \( 0.5(x^2+4x-2) = 0 \), which simplifies to \( x^2+4x-2 = 0 \). Applying the quadratic formula gives \(x = [-4 ± sqrt(16-4*1*(-2))]/2 = [-4 ± sqrt(24)]/2 = -2 ± sqrt(6)\). Hence, the x-intercepts are \( (-2+sqrt(6), 0) \) and \( (-2-sqrt(6), 0) \).
04

Check the Results Algebraically

Checking the vertex: In the standard form \( (x-h)^2 + k \), when the equation \( g(x)=0.5[(x+2)^2-6] \) has x set to -2, the equation yields -3 which matches with our calculated vertex. Checking the axis of symmetry: In the equation \( g(x)=0.5[(x+2)^2-6] \), the x-value of the vertex indeed matches to -2. Investigating the x-intercepts: Setting \( g(x) = 0 \) and solving, we already found the roots, therefore this checks out with our graphical solution.

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Key Concepts

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

Vertex of a Quadratic Function
Understanding the vertex of a quadratic function is crucial to mastering its graph. The vertex is the highest or lowest point on the graph of the quadratic function, depending on whether the parabola opens upwards or downwards.

For a quadratic function in standard form, which is expressed as \(y = ax^2+bx+c\), the vertex can be found by first converting it to the vertex form \(y=a(x-h)^2+k\), where the point (h, k) represents the vertex.

In our case, the given function \(g(x)=\frac{1}{2}(x^2+4x-2)\) is manipulated into \(g(x)=0.5[(x+2)^2-6]\) by the process of completing the square, clearly showing that the vertex of our function is at (h, k) = (-2, -3). This critical point is where the parabola changes direction, and is a fundamental piece of the puzzle for understanding the function's graph.
Axis of Symmetry
The axis of symmetry of a quadratic function is a vertical line that divides the parabola into two mirror-image halves. This line goes through the vertex, and its equation takes the form \(x = h\) where h is the x-coordinate of the vertex.

For the quadratic function \(g(x)=\frac{1}{2}(x^2+4x-2)\), after finding the vertex, we deduced that the axis of symmetry is \(x = -2\). This is a critical concept because it allows us to predict the shape and orientation of the parabola without fully graphing it. Moreover, it also aids in finding the reflection of points across the graph and better understanding the function's behavior.
X-intercepts
The x-intercepts of a quadratic function are the points where the graph crosses the x-axis, also known as the roots or zeros of the function. These can be calculated by setting the function equal to zero and solving for x.

For our function \(g(x)\), setting \(g(x) = 0\) leads to solving the equation \(x^2+4x-2 = 0\), and the quadratic formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) provides the x-intercepts. The calculations yield \(x = -2 \pm \sqrt{6}\), resulting in two x-intercepts for our parabola: \( (-2+\sqrt{6}, 0) \) and \( (-2-\sqrt{6}, 0) \). These intercepts are essential for graphing and understanding the range of possible values for x within the context of the function.
Standard Form of a Quadratic
The standard form of a quadratic function is typically expressed as \(y = ax^2+bx+c\), where a, b, and c are constants, and \(a \eq 0\). This form provides a straightforward method to analyze and graph quadratics.

From the standard form, we can determine the direction the parabola opens, the y-intercept, and the initial step in seeking the vertex and the axis of symmetry. The given function \(g(x)=\frac{1}{2}(x^2+4x-2)\) is already in standard form. By reorganizing this function into vertex form, as done in previous sections, we can uncover more about the quadratic's graph, including the crucial points of the vertex and x-intercepts, linking algebraic understanding to graphical representation.

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