## 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

1. anonymous

hmmm have you covered parabolas yet? or quadratic equations that is

2. anonymous

i believe so... and i figured out the last problem. it was -1

3. anonymous

i think its C

4. anonymous

hmm ok.. how did you get C?

5. anonymous

well im not sure. and i guarentee i dont know how to solve these

6. anonymous

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|

7. anonymous

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

8. anonymous

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

9. anonymous

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 $$\bf \textit{quadratic formula}\\ y={\color{blue}{ 8}}t^2{\color{red}{ -47}}t{\color{green}{ -2}} \qquad \qquad x= \cfrac{ - {\color{red}{ b}} \pm \sqrt { {\color{red}{ b}}^2 -4{\color{blue}{ a}}{\color{green}{ c}}}}{2{\color{blue}{ a}}}$$