Use completing the square to describe the graph of the following function support your answer graphically.
f(x)=-2x^2+4x+1
f(x)=-2x^2+4x+1
= -2(x^2 - 2x +1 - 1) + 1
= -2( (x-1)^2 -1) + 1
= -2(x+1)^2 + 3
you should be able to state the properties you are looking for from that.
To describe the graph of the function f(x) = -2x^2 + 4x + 1 using completing the square, we need to convert it into vertex form.
The vertex form of a quadratic function is given by f(x) = a(x - h)^2 + k, where (h, k) represents the coordinates of the vertex.
1. Start by taking out the common factor -2 from the first two terms: f(x) = -2(x^2 - 2x) + 1.
2. To complete the square inside the parentheses, we need to add and subtract the square of half of the coefficient of x.
i. The coefficient of x is -2, so half of it is -1.
ii. The square of -1 is 1, so we add and subtract 1 inside the parentheses: f(x) = -2(x^2 - 2x + 1 - 1) + 1.
3. Now, rewrite the equation by grouping the terms: f(x) = -2((x - 1)^2 - 1) + 1.
4. Expand and simplify: f(x) = -2(x - 1)^2 + 2 + 1.
5. Combine constants: f(x) = -2(x - 1)^2 + 3.
Now we have the equation in vertex form, f(x) = -2(x - 1)^2 + 3, which tells us that the vertex is at (1, 3).
To describe the graph of the function graphically, we can use the information about the vertex and the coefficient of x^2 to determine its shape and direction.
1. The coefficient of x^2 is -2, indicating that the parabola opens downward.
2. The vertex at (1, 3) represents the lowest point (or maximum point) on the graph.
3. Since the coefficient of x^2 is negative, the graph will be wider than the standard parabola.
4. The y-intercept is found by substituting x = 0 into the equation: f(0) = -2(0 - 1)^2 + 3 = -2(1)^2 + 3 = -2 + 3 = 1.
Putting all this information together, we can sketch the graph of the function f(x) = -2x^2 + 4x + 1. With the vertex at (1, 3), and the parabola opening downward, the graph will look like a downward-facing U-shape, with the vertex as the lowest point. The y-intercept is at (0, 1).