help_people
  • help_people
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
help_people
  • help_people
@welshfella or @triciaal or someone :D
triciaal
  • triciaal
|dw:1438815884398:dw|
zzr0ck3r
  • zzr0ck3r
I am with @triciaal

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help_people
  • help_people
ummm?
triciaal
  • triciaal
|dw:1438816139795:dw|
help_people
  • help_people
i do not understand any of this i do but what she'll i do next :)
help_people
  • help_people
what is after that part @triciaal
triciaal
  • triciaal
one approach compare your equation with the general equation identify your a, b, c and substitute
help_people
  • help_people
how she'll i do that or could you show me I'm confused if you cannot tell
triciaal
  • triciaal
alternative |dw:1438816378738:dw|
triciaal
  • triciaal
when you complete the square you are making a perfect square
triciaal
  • triciaal
|dw:1438816857649:dw|
triciaal
  • triciaal
does it make more sense now?
jdoe0001
  • 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
  • help_people
i have @jdoe0001
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
  • 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
  • jdoe0001
well... actually we took out -4, should be -6, one sec
jdoe0001
  • 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
  • jdoe0001
so... follow that so far?
help_people
  • help_people
yes
help_people
  • help_people
@jdoe0001
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
  • 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
  • help_people
-3? @jdoe0001
help_people
  • help_people
or 3 i meant to say @jdoe0001
jdoe0001
  • jdoe0001
right.... 2 * x * 3 = 6x <--- middle term
jdoe0001
  • 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
  • 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
  • help_people
wait so the final answer is a ? @jdoe0001
jdoe0001
  • jdoe0001
ahemm nope.. one sec
jdoe0001
  • 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
  • help_people
its either a or b 2 @jdoe0001
jdoe0001
  • jdoe0001
hmmm actaully
jdoe0001
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • help_people
so b? @jdoe0001
jdoe0001
  • jdoe0001
dunno... what do you think is the vertex? notice to the equation in vertex form
help_people
  • help_people
vertex is 3 and 7 @jdoe0001
help_people
  • help_people
based on that i bleive its a @jdoe0001
jdoe0001
  • jdoe0001
|dw:1438820047108:dw|
jdoe0001
  • jdoe0001
equaiont in "a" doesn't even look like the obtained equation so it can't be that
help_people
  • help_people
the only other one i would think is c @jdoe0001
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
  • jdoe0001
yeap, the maximum height of the water is 7 units vertex at 3,7|dw:1438820313064:dw|
help_people
  • help_people
so it is c?
help_people
  • help_people
@triciaal please come back he just left and i am not sure what the answer is :/
help_people
  • help_people
i first thought it was a but then he told me no so now i believe it is c
help_people
  • help_people
never mind :)

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