Answer :
Below is the solution:
f(x)=−4(x2−6x+9−9)−29
(x2−6x+9)=(x−3)(x−3)
f(x)=−4((x−3)2−9)−29
f(x)=−4(x−3)2−4(−9)−29
f(x)=−4(x−3)2+36−29
f(x=−4(x−3)2+7
f(x)=−4(x−3)2+7
That's the equation we got by completing the square.
y=a(x−h)2+k
That's the formula of a parabola in "vertex" form. Notice the similarity? The two are equal if we say
y=f(x)
a=−4
h=3
k=7
Therefore the answer is C, −4(x − 3)2 + 7; The maximum height of the water is 7 feet.
Answer:
Option 3 - [tex]y= -4(x-3)^2+7[/tex]; The maximum height of the water is 7 feet.
Step-by-step explanation:
Given : The height of water shooting from a fountain is modeled by the function [tex]f(x) = -4x^2+24x-29[/tex] where x is the distance from the spout in feet.
To find : Complete the square to determine the maximum height of the path of the water.
Solution :
[tex]f(x) = -4x^2+24x-29[/tex]
[tex]f(x) = -4(x^2-6x)-29[/tex]
Completing the square by adding and subtracting [tex](\frac{6}{2})^2=3^2=9[/tex] in the bracket,
[tex]f(x) = -4(x^2-6x+9-9)-29[/tex]
[tex]f(x) = -4((x-3)^2-9)-29[/tex]
[tex]f(x) = -4(x-3)^2+36-29[/tex]
[tex]f(x) = -4(x-3)^2+7[/tex]
The general vertex form is [tex]y=a(x-h)^2+k[/tex]
On comparing,
a=−4, h=3, k=7
The maximum height of the water is given by y-intercept i.e. k,
The maximum height of the water is 7 feet.
Therefore, Option 3 is correct.