Graph of 2x^2 - 32x + 128
In this article, we will explore the graph of the quadratic function 2x^2 - 32x + 128.
Quadratic Function
A quadratic function is a polynomial of degree two, which means the highest power of the variable (x) is two. The general form of a quadratic function is ax^2 + bx + c, where a, b, and c are constants.
Graph of 2x^2 - 32x + 128
The graph of 2x^2 - 32x + 128 is a U-shaped parabola that opens upwards. The vertex of the parabola is the lowest point on the graph.
To find the vertex of the parabola, we can use the formula:
x_vertex = -b / 2a
In this case, a = 2 and b = -32, so:
x_vertex = -(-32) / (2 * 2) x_vertex = 16 / 4 x_vertex = 4
To find the y-coordinate of the vertex, we can substitute x_vertex into the original equation:
y_vertex = 2(4)^2 - 32(4) + 128 y_vertex = 2(16) - 128 + 128 y_vertex = 32 - 128 + 128 y_vertex = 32
So, the vertex of the parabola is (4, 32).
Axis of Symmetry
The axis of symmetry is a vertical line that passes through the vertex of the parabola. In this case, the axis of symmetry is x = 4.
X-Intercepts
To find the x-intercepts, we can set the equation equal to zero and solve for x:
2x^2 - 32x + 128 = 0
Factoring the equation, we get:
(2x - 8)(x - 16) = 0
So, the x-intercepts are x = 4 and x = 16.
Y-Intercept
The y-intercept is the point where the graph crosses the y-axis. To find the y-intercept, we can substitute x = 0 into the original equation:
y = 2(0)^2 - 32(0) + 128 y = 128
So, the y-intercept is (0, 128).
Graph
Here is the graph of 2x^2 - 32x + 128:
image description: a U-shaped parabola with vertex (4, 32), x-intercepts (4, 0) and (16, 0), and y-intercept (0, 128)
In conclusion, the graph of 2x^2 - 32x + 128 is a U-shaped parabola with a vertex at (4, 32), x-intercepts at x = 4 and x = 16, and a y-intercept at (0, 128).
