Question

Sketch the graph of the quadratic function. Identify the vertex and intercepts. $$g(x)=\frac{1}{2}\left(x^{2}+4 x-2\right)$$

Short answer

The vertex of the function g(x)=1/2(x²+4x-2) is (-2,-3) and the x-intercepts are (-2+√6, 0) and (-2-√6, 0). The graph of the function is a parabola opening upwards.

Step-by-step solution

Identifying the vertex

The equation given is g(x)=1/2(x²+4x-2), which is in the standard form of g(x) = a(x-h)² + k, where the vertex of the parabola is (h,k). But it needs to be rearranged in the vertex form by completing the square method. The general method to complete the square is to take the coefficient of x, divide by 2, and then square it. Here, the coefficient of x is 4, so after performing these operations, we get \((4/2)² = 4\). Now, rewrite the given equation as \(g(x)=1/2(x²+4x+4-4-2)\), or \(g(x)=1/2((x+2)²-6)\). Hence, the vertex of the parabola is (-2,-6/2) =(-2,-3).

Identify the x-intercepts

The x-intercepts of the function are the x-values when g(x) = 0. So, solve the equation 1/2(x+2)² - 3 = 0, i.e., (x+2)² = 6. The solutions to this are \(x=-2+\sqrt{6}\) and \(x=-2-\sqrt{6}\). Hence, the x-intercepts are (-2+√6, 0) and (-2-√6, 0)

Sketch the graph

Now, using the vertex and x-intercepts sketch the graph. Plot the vertex point (-2,-3) first. Then, plot the x-intercepts (-2+√6, 0) and (-2-√6, 0). Draw the parabola making sure it passes through these points and opens upwards (since the leading coefficient 1/2 is positive).

Key concepts

Completing the Square

Completing the square is a technique used to rewrite quadratic functions in a form that more clearly shows certain properties, like the vertex of the parabola. Let's look at how we can use this method to solve for the vertex in our provided exercise.

When we have a function in the format of \( ax^2 + bx + c \), we can complete the square by transforming it into the vertex form, which is \( a(x-h)^2 + k \). This format immediately reveals the vertex as the point \( (h, k) \). To do this, we follow a few steps: First, we make sure the coefficient of \( x^2 \) is 1, if not we factor it out. Next, we focus on the \( bx \) term, here the coefficient is 'b'. We take \( b/2 \), square it, and add it inside the bracket but ensure to subtract the same value to maintain equality. In the exercise, the coefficient of \( x \) is 4, so we calculate \( (4/2)^2 = 4 \). We then rewrite the given quadratic function inserting \( (x+2)^2 \) into the equation and adjusting so that the equation remains balanced.

The beauty of completing the square is that, even if it sounds complex, once you grasp it, the process is a repeatable, step-by-step approach that provides a clear path to finding important characteristics of the quadratic function, such as its vertex.

Vertex of a Parabola

The vertex of a parabola is the 'tip' of the curve, where it changes direction. For the upward or downward opening parabolas, the vertex represents the maximum or minimum point of the function, respectively. To understand the vertex in the context of our exercise, we leveraged the completed square form of the quadratic function.

From the vertex form \( g(x) = a(x-h)^2 + k \), we can directly read off the vertex as the point \( (h, k) \). In our example, after completing the square, we determined the vertex form to be \( g(x) = \frac{1}{2}(x+2)^2 - 3 \), revealing the vertex to be at the coordinates \( (-2, -3) \).

The significance of identifying the vertex lies in its ability to offer a reference point for sketching the entire parabola and understanding the function's behavior. It is vital for understanding both the graph's shape and location in the coordinate plane.

X-Intercepts

X-intercepts are the points where the graph of a function crosses the x-axis, indicating the values of 'x' for which the function \( g(x) \) equals zero. To find them, as done in the exercise, you must solve the equation \( g(x)=0 \).

Using the vertex form we obtained earlier, \( \frac{1}{2}(x+2)^2 - 3 = 0 \), we set the expression inside the brackets equal to 6 to cancel out the subtraction of 3, after which we find the square root to release the \( x \) from the binomial square. This provides two solutions: \( x = -2 + \sqrt{6} \) and \( x = -2 - \sqrt{6} \), which gives us the x-intercepts for our parabola. These are the points \( (-2 + \sqrt{6}, 0) \) and \( (-2 - \sqrt{6}, 0) \).

Understanding x-intercepts is crucial for graphing and analyzing real-world situations since these points often represent meaningful occurrences, such as the time when a projectile hits the ground in physics, or break-even points in economics.

Parabolic Graph Sketching

Sketching the graph of a parabola entails plotting key features such as the vertex and x-intercepts and then drawing a smooth curve through these points. Begin by plotting the vertex, the highest or lowest point on the graph, which acts as a turning point. In our case, the vertex is \( (-2, -3) \).

Once the vertex is in place, plot the x-intercepts we found earlier, which are \( (-2 + \sqrt{6}, 0) \) and \( (-2 - \sqrt{6}, 0) \). With these points marked, you can draw a symmetrical curve that represents the parabola, ensuring that it is opened upward because the leading coefficient \( \frac{1}{2} \) is positive.

It's important to note that while the x-intercepts and the vertex give us vital points on the graph, the line of symmetry which passes through the vertex and divides the parabola into two mirrored halves, is also an important feature. For precision, additional values can be calculated and plotted. Parabolic graph sketching is a fundamental skill in many fields of study, including physics, engineering, and finance, due to the frequent appearance of quadratic relationships.