91Ó°ÊÓ

Quadratic functions can be presented in several forms, and one of the most insightful is the vertex form. This form is expressed as \(f(x) = a(x - h)^2 + k\), where \(a\), \(h\), and \(k\) are constants. The values \(h\) and \(k\) provide the coordinates of the vertex of the parabola, which is the highest or lowest point depending on whether the parabola opens upwards or downwards.

The function \(g(x) = 2(x+1)^2 + 2\) is already in vertex form. By comparing this with the standard vertex form \(f(x) = a(x - h)^2 + k\), you can identify the vertex as \((-1, 2)\).

This indicates that the parabola reaches its minimum at the point \((-1, 2)\) since the coefficient \(a = 2\) is positive, causing the parabola to open upwards. The vertex form is particularly useful for identifying the transformation of the graph from the parent function \(y = x^2\). In this specific case, the parabola is shifted one unit to the left and two units up.

To find the x-intercepts of a quadratic function, you solve the equation when the output \(g(x)\) equals zero. This process involves identifying the x-values for which the quadratic function intersects the x-axis, meaning the y-value is zero.

For the function \(g(x) = 2(x+1)^2 + 2\), setting it to zero results in the equation \(0 = 2(x + 1)^2 + 2\). Simplifying the equation, you first subtract 2 from both sides to get \(-2 = 2(x + 1)^2\), and then divide by 2, leading to \(-1 = (x + 1)^2\).

Since we know that \((x+1)^2\) is a squared term, it can never be negative. This implies that there are no real numbers that satisfy this equation, indicating there are no x-intercepts for this function. If a parabola does not cross or touch the x-axis, it has complex roots instead, pointing here to an imaginary component.

Finding the y-intercept of a quadratic function is straightforward. You simply evaluate the function at \(x = 0\), revealing the point at which the curve intersects the y-axis. This is where the x-value of the function is zero.

For the function \(g(x) = 2(x+1)^2 + 2\), substitute \(x = 0\):
\[g(0) = 2(0 + 1)^2 + 2 = 2(1) + 2 = 4\]
Thus, the y-intercept is the point \((0, 4)\). This shows that the graph of the function crosses the y-axis at the value of 4. The y-intercept provides a useful starting point for graphing the quadratic and understanding its relation to the rest of the coordinate plane.