y=x^2 -4x -32. Use this function and identify the axis of symmetry and the vertex.
Thank you.
By completing the square,
y=x^2-4x-32
=(x-2)²-4-32
=(x-2)²-36
The line x-2=0 passes through the vertex, hence the vertex is
(2,y(2))=(2, -36)
The line of symmetry of a parabola that opens up or down is the vertical line through the vertex.
To identify the axis of symmetry and the vertex of the function y = x^2 - 4x - 32, you can use the standard form of a quadratic equation, which is y = ax^2 + bx + c.
In this case, a = 1, b = -4, and c = -32.
The axis of symmetry for a quadratic equation is given by the formula x = -b / (2a).
First, plug in the values of a and b into the formula:
x = -(-4) / (2 * 1) = 4 / 2 = 2
So, the x-coordinate of the vertex and the axis of symmetry is x = 2.
Next, substitute this value back into the original equation to find the y-coordinate.
y = (2)^2 - 4(2) - 32
y = 4 - 8 - 32
y = -36
Therefore, the vertex is located at the point (2, -36).
The axis of symmetry, where the parabola is symmetrical, is a vertical line passing through the vertex, which is x = 2. And the vertex is the lowest point (in this case, a maximum since the coefficient of x^2 is positive) of the parabola.