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anonymous
 one year ago
Someone help please!!!
A basketball is thrown upwards. The height f(t), in feet, of the basketball at different times t, in seconds, is shown by the following function:
f(t) = 16t2 + 94t + 12
Which of the following is a reasonable domain of the graph of the function when the basketball falls from its highest height to the ground? (4 points)
3 ≤ x ≤ 6
2 ≤ x ≤ 5
1 ≤ x ≤ 4
0 ≤ x ≤ 3
anonymous
 one year ago
Someone help please!!! A basketball is thrown upwards. The height f(t), in feet, of the basketball at different times t, in seconds, is shown by the following function: f(t) = 16t2 + 94t + 12 Which of the following is a reasonable domain of the graph of the function when the basketball falls from its highest height to the ground? (4 points) 3 ≤ x ≤ 6 2 ≤ x ≤ 5 1 ≤ x ≤ 4 0 ≤ x ≤ 3

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jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.0hmmm have you covered parabolas yet? or quadratic equations that is

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i believe so... and i figured out the last problem. it was 1

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.0hmm ok.. how did you get C?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well im not sure. and i guarentee i dont know how to solve these

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.0hmm ok well let us notice first off, is a quadratic equation so its graph is a parabola the leading term coefficient is negative meaning the parabola is opening downwardsdw:1438296963025:dw

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.0what you're being asked is what is a reasonable domain, or values for "x" between the "peek" or Uturn and the time it hits the grouns notice the graph, the object goes UP then reaches a peak and makes a Uturn and it goes DOWN and hits the ground dw:1438297081023:dw that is the domain you're being asked

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.0so... you'd want to, for that, solve the equation for "x", by either factoring or using the quadratic formula so you can find the xintercept, or when it hits the ground and you'd want to find the vertex, the vertex is simple \(\bf \textit{vertex of a parabola}\\ \quad \\ y = {\color{red}{ 16}}t^2{\color{blue}{ +94}}t{\color{green}{ +12}}\qquad \left(\cfrac{{\color{blue}{ b}}}{2{\color{red}{ a}}}\quad ,\quad {\color{green}{ c}}\cfrac{{\color{blue}{ b}}^2}{4{\color{red}{ a}}}\right)\) now, to find the xintercepts, or the roots

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.0anyhow to get the roots you could use say \(\bf f(t)=16t^2+94t+12\implies 0=16t^2+94t+12 \\ \quad \\ 16t^294t12=0\implies 8t^247t6=0\) and that one, doesn't factor out nicely thus you'd need to use the quadratic formula that is \(\bf \textit{quadratic formula}\\ y={\color{blue}{ 8}}t^2{\color{red}{ 47}}t{\color{green}{ 2}} \qquad \qquad x= \cfrac{  {\color{red}{ b}} \pm \sqrt { {\color{red}{ b}}^2 4{\color{blue}{ a}}{\color{green}{ c}}}}{2{\color{blue}{ a}}}\)
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