To find the y-intercept, set x = 0 in the equation y = 64 - x^2:
y = 64 - 0^2
y = 64
So, the y-intercept is at the point (0, 64).
To find the x-intercepts, set y = 0 in the equation y = 64 - x^2:
0 = 64 - x^2
x^2 = 64
x = ±√64
x = ±8
So, the x-intercepts are at the points (-8, 0) and (8, 0).
To find the coordinates of the vertex, we need to find the x-coordinate of the vertex using the formula x = -b/(2a) where a is the coefficient of x^2 and b is the coefficient of x in the equation y = ax^2 + bx + c:
In this case, a = -1, b = 0, and c = 64.
x = -0 / (2 * -1)
x = 0
Now, substitute x = 0 into the equation y = 64 - x^2 to find the y-coordinate of the vertex:
y = 64 - 0^2
y = 64
So, the coordinates of the vertex are (0, 64).
The equation of the axis of symmetry is x = 0.
Since the coefficient of x^2 is negative, the parabola opens downwards, so the vertex is the maximum point of the parabola.
y=64-x^2.
y-intercept?
X-intercept(s)?
Coordinates of the vertex?
Equation of axis of symmetry?
Maximum or minimum?
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