Respuesta :
The given function is [tex]f(x) = -x^2 + 50x - 264[/tex]. This is a quadratic function because the degree of the polynomial, the variable with the highest exponent, is 2.
All quadratic functions can be expressed in the form [tex]f(x) = ax^2 + bx + c[/tex].
In your equation,
a = -1
b = 50
c = -264
The equation for the x-coordinate vertex of a quadratic function is [tex]x = \frac{-b}{2a} [/tex].
Applying this to the given function,
[tex]x = \frac{-(50)}{2(-1)} = \frac{50}{2} = 25[/tex].
Now plug the value of x into f(x) to find f(x) (aka the y-coordinate).
[tex]f(x) = -x^2 + 50x - 264 \\ f(25) = -(25)^2 + 50(25) - 264 \\ f(25) = 361[/tex]
The vertex is (25, 361)
Part B
The x-intercept is where the graph crosses the x-axis. Therefore, f(x) (aka the y-coordinate) will be 0. So the x-intercept will be in the form of (x, 0).
To find the x-intercept, set f(x) = 0.
[tex]0 = -x^2 + 50x - 264 \\ 0 = (x-6)(x-44) \\ x = 6, x = 44[/tex].
Therefore the x-intercepts will be at (6, 0) and (44, 0)
All quadratic functions can be expressed in the form [tex]f(x) = ax^2 + bx + c[/tex].
In your equation,
a = -1
b = 50
c = -264
The equation for the x-coordinate vertex of a quadratic function is [tex]x = \frac{-b}{2a} [/tex].
Applying this to the given function,
[tex]x = \frac{-(50)}{2(-1)} = \frac{50}{2} = 25[/tex].
Now plug the value of x into f(x) to find f(x) (aka the y-coordinate).
[tex]f(x) = -x^2 + 50x - 264 \\ f(25) = -(25)^2 + 50(25) - 264 \\ f(25) = 361[/tex]
The vertex is (25, 361)
Part B
The x-intercept is where the graph crosses the x-axis. Therefore, f(x) (aka the y-coordinate) will be 0. So the x-intercept will be in the form of (x, 0).
To find the x-intercept, set f(x) = 0.
[tex]0 = -x^2 + 50x - 264 \\ 0 = (x-6)(x-44) \\ x = 6, x = 44[/tex].
Therefore the x-intercepts will be at (6, 0) and (44, 0)
Answer:
The vertex is (25,361), the x intercepts are (6,0) and (44,0).
Step-by-step explanation:
Consider the provided function,
[tex]f(x) = -x^2 + 50x - 264 [/tex]
The standard equation of the quadratic equation is [tex]f(x) = ax^2 + bx + c[/tex]
And the vertex of the quadratic equation can be calculated as:
[tex]h=\frac{-b}{2a}[/tex] and k = f(h)
Part A:
By comparing the provided equation with the standard equation we can concluded that a = -1, b = 50 and c = -264
Now use the vertex formula and substitute the respective value in the formula that will give us the x coordinate of the vertex.
[tex]h=\frac{-50}{2(-1)}[/tex]
[tex]h=\frac{50}{2}[/tex]
[tex]h=25[/tex]
Now substitute h = 25 in the provided equation as shown:
[tex]f(h) = -(25)^2 + 50(25) - 264 [/tex]
[tex]f(h) = -625 + 1250 - 264 [/tex]
[tex]f(h) =361 [/tex]
Hence, the vertex is (25,361).
The provided equation has a negative x² coefficient that means it is a downward parabola having maximum at vertex (25,361) but there exist no minimum point.
In context of the problem: The profit will be maximum when the number of membership sold is 25.
Part B:
To determine the x intercept simply substitute f(x)=0 and solve for x.
[tex]-x^2 + 50x - 264=0[/tex]
[tex]x^2 -50x + 264=0[/tex]
[tex]x^2 -6x-44x +264=0[/tex]
[tex]x(x-6)-44(x-6)=0[/tex]
[tex](x-6)(x-44)=0[/tex]
[tex]x=6\ or\ x=44[/tex]
Hence, the x intercepts are (6,0) and (44,0).
That means, there will be zero profit when the number of memberships sold are 6 or 44.
