91Ó°ÊÓ

1. For the first function, \(y = x^2\), we can tell that it's a simple quadratic function (a parabola) with the vertex at the origin \((0,0)\) and opens upwards because the leading coefficient (1) is positive. 2. For the second function, \(y = 2x^2 - 4x + 1\), we see that it's also a quadratic function, and it will also have a parabolic shape with its vertex shifted from the origin. We will find the vertex later to guide us in the transformation process.

To sketch the graph of \(y=x^2\), we can plot a few points. Consider the points (-1,1), (0,0), and (1,1) on the graph. These points show a basic upward-opening parabola.

To find the vertex of the second function, which is in the form \(y = ax^2+bx+c\), we can use the formula for the vertex, \((h,k)\), given by: \(h = \frac{-b}{2a} \) and \(k = c - \frac{b^2}{4a}\) For the given function, \(y=2x^2-4x+1\), we have \(a = 2\), \(b = -4\), and \(c = 1\). Plugging these values into the vertex formula gives: \(h = \frac{-(-4)}{2(2)} = \frac{4}{4} = 1 \) \(k = 1 - \frac{(-4)^2}{4(2)} = 1 - 4 = -3\) So, the vertex of the second function is \((1,-3)\).

Now that we have the vertex of the second function, we can use transformations to obtain its graph from the sketch of the first function, \(y=x^2\). Notice that the second function has a leading coefficient of 2, which will make the parabola narrower and steeper. Therefore, we can apply the following transformations: 1. Stretch the parabola vertically by a factor of 2: \(y = 2x^2\) 2. Translate the parabola left by one unit: \(y = 2(x - 1)^2\) 3. Translate the parabola down by three units: \(y = 2(x - 1)^2 - 3\) At this point, you should have the second function's graph, \(y = 2x^2-4x+1\), which is a steeper parabola with vertex (1,-3). By following these steps and applying the transformations, you can sketch the graph of the given functions.