anonymous
  • anonymous
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
Mathematics
chestercat
  • chestercat
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jdoe0001
  • jdoe0001
hmmm have you covered parabolas yet? or quadratic equations that is
anonymous
  • anonymous
i believe so... and i figured out the last problem. it was -1
anonymous
  • anonymous
i think its C

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jdoe0001
  • jdoe0001
hmm ok.. how did you get C?
anonymous
  • anonymous
well im not sure. and i guarentee i dont know how to solve these
jdoe0001
  • jdoe0001
hmm 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 downwards|dw:1438296963025:dw|
jdoe0001
  • jdoe0001
what you're being asked is what is a reasonable domain, or values for "x" between the "peek" or U-turn and the time it hits the grouns notice the graph, the object goes UP then reaches a peak and makes a U-turn and it goes DOWN and hits the ground |dw:1438297081023:dw| that is the domain you're being asked
jdoe0001
  • jdoe0001
so... you'd want to, for that, solve the equation for "x", by either factoring or using the quadratic formula so you can find the x-intercept, 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 x-intercepts, or the roots
jdoe0001
  • jdoe0001
anyhow to get the roots you could use say \(\bf f(t)=-16t^2+94t+12\implies 0=-16t^2+94t+12 \\ \quad \\ 16t^2-94t-12=0\implies 8t^2-47t-6=0\) 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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