Question

\(g(x)=2(x-1)^2+2\)

Short answer

The vertex of the function \(g(x)=2(x-1)^2+2\) is at the point \((1,2)\). The parabola opens upwards and is narrower than \(y=x^2\). The function crosses the y-axis at \(y=4\).

Step-by-step solution

Identify the Vertex

The vertex of a quadratic function in standard form \(f(x)=a(x-h)^2+k\) is given by the point \((h,k)\). For the function \(g(x)=2(x-1)^2+2\), the vertex is at point \((1,2)\).

Identify Shape and Direction

The parabola's shape and direction are determined by the coefficient of \(x\), which in this case is '2'. Since '2' is a positive number, the parabola opens upwards, and the factor of '2' indicates that it is narrower than the standard parabola \(y=x^2\).

Find Y-intercept

The y-intercept is found by setting \(x = 0\) in the equation of the function and solving for \(y\). Therefore, if we set \(x = 0\) in \(g(x)=2(x-1)^2+2\), we get \(y = 2\cdot(0-1)^2+2 = 4\). So, the y-intercept of the function is at \(y = 4\).

Plot the Function

With the above information, start plotting the function. Plot the vertex at \((1, 2)\). The parabola is narrow and opens upwards, and crosses the y-axis at \((0, 4)\). Plot a few more points if necessary and sketch the parabola.

Key concepts

Vertex of a Parabola

The vertex of a parabola is a crucial concept in understanding the graph of a quadratic function. Essentially, it's the highest or lowest point on the curve, acting as a pivot around which the parabola is symmetric.

In the case of the quadratic function given by the equation \(g(x)=2(x-1)^2+2\), the vertex can be directly read off as \(h\) and \(k\) from the general vertex form \(f(x)=a(x-h)^2+k\), leading to the point \( (1, 2) \). This point is central to the graph and will be your starting place when sketching the curve.

To understand why it's called a 'vertex', you can imagine it as the peak of a mountain or the bottom of a valley, representing the maximum or minimum of the quadratic function.

Parabola Shape and Direction

A parabola's shape and direction reveal a lot about the quadratic function it represents. These characteristics are decided by the coefficient of the \(x^2\) term, which is the 'a' value in the vertex form.

With the function \(g(x)=2(x-1)^2+2\), the coefficient is '2'. Since this is a positive value, the parabola opens upwards, like a traditional 'U' shape. If this coefficient were negative, the parabola would open downwards, resembling an upside-down 'U'. Moreover, the magnitude of this coefficient affects how wide or narrow the parabola is when compared to the standard parabola \(y=x^2\). Here, because we have a coefficient of '2', this means that our parabola is narrower than the standard one.

It's fascinating to see just how much the 'a' value shapes the graph. It determines not just whether you're graphing a 'smiley' or a 'frowny', but also how slim or broad the smile or frown appears.

Y-Intercept

Another fundamental concept is the y-intercept of the parabola, which is the point where the parabola crosses the y-axis. To find this value, we set \(x = 0\) in our quadratic function and solve for \(y\).

For the function \(g(x)=2(x-1)^2+2\), when \(x = 0\), we calculate the y-coordinate to be \(4\), giving us the y-intercept at the point \( (0, 4) \). The y-intercept is significant not only for graphing purposes, but it can also represent the initial value in real-world applications, such as the starting height of an object that's been launched upward.

Remember, finding the y-intercept is typically straightforward in quadratic functions since it involves zero input for \(x\), and therefore simplifies the equation. It serves as a key checkpoint on the graph, alongside the vertex, for sketching the entire curve accurately.