The height of water shooting from a fountain is modeled by the function f(x) = −4x2 + 24x − 29 where x is the distance from the spout in feet. Complete the square to determine the maximum height of the path of the water.
−4(x − 3)2 − 29; The maximum height of the water is 3 feet.
−4(x − 3)2 − 29; The maximum height of the water is 29 feet.
−4(x − 3)2 + 7; The maximum height of the water is 7 feet.
−4(x − 3)2 + 7; The maximum height of the water is 3 feet.

- help_people

- Stacey Warren - Expert brainly.com

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- help_people

@welshfella or @triciaal or someone :D

- triciaal

|dw:1438815884398:dw|

- zzr0ck3r

I am with @triciaal

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## More answers

- help_people

ummm?

- triciaal

|dw:1438816139795:dw|

- help_people

i do not understand any of this i do but what she'll i do next :)

- help_people

what is after that part @triciaal

- triciaal

one approach
compare your equation with the general equation
identify your a, b, c and substitute

- help_people

how she'll i do that or could you show me I'm confused if you cannot tell

- triciaal

alternative |dw:1438816378738:dw|

- triciaal

when you complete the square you are making a perfect square

- triciaal

|dw:1438816857649:dw|

- triciaal

does it make more sense now?

- jdoe0001

@help_people have you covered what a "perfect square trinomial" is?
sometimes just called "perfect square"
if not, this may be the time to check your book for it then

- help_people

i have @jdoe0001

- jdoe0001

so.. this rings a bell \(\begin{array}{cccccllllll}
{\color{brown}{ a}}^2& + &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\
\downarrow && &&\downarrow \\
{\color{brown}{ a}}&& &&{\color{blue}{ b}}\\
&\to &({\color{brown}{ a}} + {\color{blue}{ b}})^2&\leftarrow
\end{array}\qquad
% perfect square trinomial, negative middle term
\begin{array}{cccccllllll}
{\color{brown}{ a}}^2& - &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\
\downarrow && &&\downarrow \\
{\color{brown}{ a}}&& &&{\color{blue}{ b}}\\
&\to &({\color{brown}{ a}} - {\color{blue}{ b}})^2&\leftarrow
\end{array}\)
right?

- jdoe0001

let's see your equation, and let us do some grouping, as triciaal was indicating
\(\bf f(x)=-4x^2+24x-29\implies y=-4x^2+24x-29
\\ \quad \\
y=(-4x^2+24x)-29\implies y=-4(x^2+6x)-29\impliedby \textit{common factor}\)
follow that?

- jdoe0001

well... actually we took out -4, should be -6, one sec

- jdoe0001

\(\bf f(x)=-4x^2+24x-29\implies y=-4x^2+24x-29
\\ \quad \\
y=(-4x^2+24x)-29\implies y=-4(x^2-6x)-29\impliedby \textit{common factor}\)

- jdoe0001

so... follow that so far?

- help_people

yes

- help_people

@jdoe0001

- jdoe0001

ok
so \(\bf y=-4(x^2-6x)-29\implies y=-4(x^2-6x+{\color{red}{ \square }}^2)-29\)
so... we have a missing number there, to make a "perfect square trinomial"
what do you think that number is anyway?

- jdoe0001

notice \(\begin{array}{cccccllllll}
{\color{brown}{ a}}^2& + &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\
\downarrow && &&\downarrow \\
{\color{brown}{ a}}&& &&{\color{blue}{ b}}\\
&\to &({\color{brown}{ a}} + {\color{blue}{ b}})^2&\leftarrow
\end{array}\qquad
% perfect square trinomial, negative middle term
\begin{array}{cccccllllll}
{\color{brown}{ a}}^2& - &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\
\downarrow && &&\downarrow \\
{\color{brown}{ a}}&& &&{\color{blue}{ b}}\\
&\to &({\color{brown}{ a}} - {\color{blue}{ b}})^2&\leftarrow
\end{array}\)
the middle term is just
the other two terms without the "2" exponent
times 2
and our last number is hidden in the middle term

- help_people

-3? @jdoe0001

- help_people

or 3 i meant to say @jdoe0001

- jdoe0001

right.... 2 * x * 3 = 6x <--- middle term

- jdoe0001

bear in mind two things
one
there's a -4 outside the group, that is multiplying the group
and that
what we're really doing is borrowing from our good fellow Mr Zero, 0
so, if we add anything, we also have to subtract it
so, we'll add \(3^2\) but is being multiplied by the -4 outside, that means that we're really adding \(3^2 \)cdot -4 = -36

- jdoe0001

and if we're adding -36, well, really subtracting, since it's negative
we also have to add it
so that -36 + 36 = 0 and our good felow Mr Zero gets back his figure

- help_people

wait so the final answer is a ? @jdoe0001

- jdoe0001

ahemm nope.. one sec

- jdoe0001

\(\bf y=-4(x^2-6x)-29\implies y=-4(x^2-6x+{\color{red}{ 3 }}^2)+36-29
\\ \quad \\
y=-4(x-3)^2-7
\\ \quad \\ \quad \\
y=(x-{\color{brown}{ h}})^2+{\color{blue}{ k}}
\qquad\qquad vertex\ ({\color{brown}{ h}},{\color{blue}{ k}})\)
so... what do you think is the vertex? or that y-coordinate for that matter

- help_people

its either a or b 2 @jdoe0001

- jdoe0001

hmmm actaully

- jdoe0001

\(\bf yy=-4(x^2-6x)-29\implies y=-4(x^2-6x+{\color{red}{ 3 }}^2)+36-29
\\ \quad \\
y=-4(x-{\color{brown}{ 3}})^2+{\color{blue}{ 7}}
\\ \quad \\ \quad \\
y=(x-{\color{brown}{ h}})^2+{\color{blue}{ k}}
\qquad\qquad vertex\ ({\color{brown}{ h}},{\color{blue}{ k}})
\)
can you see the vertex now?

- jdoe0001

shoot even got 2 yy, =(
\(\bf y=-4(x^2-6x)-29\implies y=-4(x^2-6x+{\color{red}{ 3 }}^2)+36-29
\\ \quad \\
y=-4(x-{\color{brown}{ 3}})^2+{\color{blue}{ 7}}
\\ \quad \\ \quad \\
y=(x-{\color{brown}{ h}})^2+{\color{blue}{ k}}
\qquad\qquad vertex\ ({\color{brown}{ h}},{\color{blue}{ k}})
\)

- jdoe0001

see the vertex now? see the y-coordinate?
the y-coordinate of the verex, is the "maximum height"
because the vertex is the U-turn point for the graph, it goes up, reaches a maximum, then back down

- help_people

ok so i am trying to find the answer lol is it a or b i said it was a but you said i was wrong @jdoe0001

- jdoe0001

well.... thata form of the quadratic is the so-called "vertex form"
because it shows the vertex coordinates
"x" is the horizontal distance
"y" is the vertical height
the equation has a leading term negative coefficient, thus is opening downwards
thus the maximum point for it, is at the vertex

- jdoe0001

and is obtained by "completing the square"
so... the vertex should be pretty obvious there
and so is the y-coordinate of the vertex
and that y-coordinate, is the maximum height of it

- help_people

so b? @jdoe0001

- jdoe0001

dunno... what do you think is the vertex? notice to the equation in vertex form

- help_people

vertex is 3 and 7 @jdoe0001

- help_people

based on that i bleive its a @jdoe0001

- jdoe0001

|dw:1438820047108:dw|

- jdoe0001

equaiont in "a" doesn't even look like the obtained equation
so it can't be that

- help_people

the only other one i would think is c @jdoe0001

- jdoe0001

\(\bf -4(x - {\color{brown}{ 3}})^2 {\color{blue}{ -29 }}\)
according to that, the vertex is 3, -29
which is not what the simplified version is

- jdoe0001

yeap, the maximum height of the water is 7 units
vertex at 3,7|dw:1438820313064:dw|

- help_people

so it is c?

- help_people

@triciaal please come back he just left and i am not sure what the answer is :/

- help_people

i first thought it was a but then he told me no so now i believe it is c

- help_people

never mind :)

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