How do you identify the vertex of y= xsquared + 16x+64
y = x^2+16x+64
= (x+8)^2 + 0
so, the vertex is at (-8,0)
To identify the vertex of the quadratic equation y = x^2 + 16x + 64, you can use the formula -b/2a.
In the given equation, a = 1 (coefficient of x^2) and b = 16 (coefficient of x).
Using the formula, the x-coordinate of the vertex is -b/2a = -16/(2*1) = -16/2 = -8.
To find the y-coordinate of the vertex, substitute the x-coordinate (-8) into the equation:
y = (-8)^2 + 16(-8) + 64 = 64 - 128 + 64 = 0.
Therefore, the vertex of the quadratic equation y = x^2 + 16x + 64 is (-8, 0).
To identify the vertex of the quadratic function y = x^2 + 16x + 64, you can use a formula called the vertex formula. The vertex formula provides the x-coordinate of the vertex of a quadratic function in the form y = ax^2 + bx + c.
The vertex formula is given by:
x = -b / (2a)
In this case, the quadratic function is y = x^2 + 16x + 64, which means a = 1, b = 16, and c = 64. Plugging these values into the vertex formula, we get:
x = -16 / (2 * 1)
x = -16 / 2
x = -8
So, the x-coordinate of the vertex is -8.
To find the y-coordinate of the vertex, you can substitute the value of x into the original quadratic function:
y = (-8)^2 + 16(-8) + 64
y = 64 - 128 + 64
y = 0
Therefore, the vertex of the quadratic function y = x^2 + 16x + 64 is (-8, 0).