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Chapter 2: Problem 35

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

Short Answer

Expert verified
The given function describes a parabola that opens upwards and is more narrow than the standard parabola. The vertex is located at (1, -2) and the y-intercept is at y = 0.

Step by step solution

01

Identify the vertex

The vertex of the function can be found by identifying the values of 'h' and 'k' from the given function. In our function \(g(x) = 2(x - 1)^2 - 2\), this means h = 1, and k = -2. So the vertex of our parabola is at point (1, -2).
02

Identify the direction and width of the parabola

This is found by analyzing the 'a' coefficient. If 'a' is positive, the parabola opens upward, if 'a' is negative, the parabola opens downward. Also, the higher the value of 'a' (ignoring the sign), the narrower the parabola. Given that 'a' = 2 for our function, it indicates that our parabola opens upward and is twice as narrow compared to the standard parabola \((y = x^2)\)
03

Calculate the y-intercept

To find the y-intercept, set 'x' equal to 0 in your function and solve for 'g(x)'. When x=0, \(g(x) = 2(0-1)^2 - 2 = 0\), so the y-intercept is at y=0.

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Key Concepts

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

Vertex of a Parabola
Every parabola has a vertex, which is the highest or lowest point on the graph, and is crucial for understanding its shape. In the quadratic function g(x) = 2(x - 1)^2 - 2, the vertex can be found easily because the function is in vertex form, y = a(x - h)^2 + k, where (h, k) represents the vertex coordinates.

In this case, h = 1, and k = -2, indicating that the vertex of the parabola g(x) is at (1, -2). This point is a cornerstone of the graph, serving as a reference for all other points and helping to visualize the parabola's path. By identifying the vertex, one can easily sketch the parabola's axis of symmetry, which is a vertical line running through the vertex, specifically at x = h in this instance.
Direction of a Parabola
The direction in which a parabola opens, upward or downward, is determined by the coefficient a in the quadratic function. For the given function g(x) = 2(x - 1)^2 - 2, we have a = 2, which is positive. This positive value tells us that the parabola opens upward.

Additionally, the value of a also affects the width or 'steepness' of the parabola. The larger the absolute value of a, the narrower the parabola. Since our a is 2, the parabola is narrower compared to the parabola of the basic function y = x^2, which has a = 1. A larger a means the sides of the parabola rise more steeply from the vertex, while a smaller positive a would indicate a wider, more gently sloping parabola.
Y-Intercept Calculation
The y-intercept is where the parabola crosses the y-axis. To find this point, set x = 0 in the equation and solve for y. For the equation g(x) = 2(x - 1)^2 - 2, when we replace x with 0, the calculation is g(0) = 2(0 - 1)^2 - 2. Simplifying this gives us g(0) = 2(1) - 2, which equals 0. Thus, the y-intercept is (0, 0).

The y-intercept is a significant point on the graph because it provides another reference point alongside the vertex. For any parabola, there鈥檚 always only one y-intercept, making it a unifying characteristic for all quadratic functions. Knowing the y-intercept allows students to start plotting the graph with two main points: the vertex and the y-intercept.

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Most popular questions from this chapter

\(\frac{5}{2}=\frac{-20}{x}\)

\(f(x)=x^2\); vertical stretch by a factor of 4 and a reflection in the \(x\)-axis, followed by a translation 2 units up

\(y=6 x^2-1\)

ANALYZING EQUATIONS The graph of which function has the same axis of symmetry as the graph of \(y=x^2+2 x+2 ?\) (A) \(y=2 x^2+2 x+2\) (B) \(y=-3 x^2-6 x+2\) (C) \(y=x^2-2 x+2\) (D) \(y=-5 x^2+10 x+2\)

\(y=(x-2)^2-1\)
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