find the x-intercepts, f(x)=-9x^2-48x-64
f(x) = -1 (3x + 8)^2
the graph is tangent to the x-axis
... at (-8/3,0)
To find the x-intercepts of a function, we need to set the value of f(x) equal to zero and solve for x. The x-intercepts are the points where the graph of the function intersects the x-axis.
Given f(x) = -9x^2 - 48x - 64, we can set f(x) equal to zero:
-9x^2 - 48x - 64 = 0
Now, let's solve for x using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -9, b = -48, and c = -64.
Plugging these values into the quadratic formula:
x = (-(-48) ± √((-48)^2 - 4(-9)(-64))) / (2(-9))
x = (48 ± √(2304 - 2304)) / (-18)
x = (48 ± √0) / (-18)
Since the discriminant (√(b^2 - 4ac)) is zero, the quadratic equation has a double root.
Simplifying further:
x = (48 ± 0) / (-18)
x = 48 / (-18)
x = -8/3
Therefore, the function f(x) = -9x^2 - 48x - 64 has a double x-intercept at x = -8/3.