The equation h=−16t2+100t+5 represents the height, in feet, (h) of the arrow after t seconds elapsed. The question is essentially asking after how seconds (t) will the arrow reach a height of 20 feet (h).
20=−16r2+100t+5 | Subtract 20 from both sides. |
−16t2+100t−15=0 | Use the quadratic formula to find the solutions of this equation. |
t=−100±√1002−4(−16)(−15)2(−16) | Now, simplify from here. |
t=−100±√9040−32 | 16 is the largest perfect square factor of 9040, so we can simplify the radical further. |
√9040=√16∗565=√16√565=4√565 | |
t=−100±4√565−32 | -100,4, and -32 all have a GCF of 4, so this answer can still be simplified further. |
t=−25±√565−8 | |
t1=6.096sect2=0.154 | Since the question specified that the arrow had to reach the height only after reaching its maximum altitude, then the only correct answer is 6.096 seconds. |
If I am not mistaken, this problem requires you to generate the cubic equation of the four points listed. I see that f(0)=0, so one of the coordinates of the x-intercept is certainly located at (0,0).I also notice that there must be another zero located at \(1 because the graph crosses the x-axis there, but we cannot determine exactly where it is yet. An odd-degree polynomial cannot have an even number of real zeros (unless a multiplicity of a zero is even). We have determined already that there are 2 zeros, so a third must be lying somewhere. Yet again, we cannot determine where.
Cubic functions are written in the form f(x)=ax3+bx2+cx+d. We can actually solve for d straightaway by plugging in f(0).
f(x)=ax3+bx2+cx+d | Plug in f(0) |
f(0)=a∗03+b∗02+c∗0+d | There is a lot of simplification that can occur here! |
f(0)=d | According to your original three points, f(0)=0, so substitute that in. |
0=d | |
Great! We have already solved for one of the missing variables. Our original equation is now f(x)=ax3+bx2+cx. It only gets harder from here, though. Let's plug in the remaining points, and let's see what happens!
f(x)=ax3+bx2+cx | Evaluate this function at f(-1). |
f(−1)=a(−1)3+b(−1)2+c(−1) | Ok, now simplify. |
f(−1)=−a+b−c | By definition, f(−1)=15. |
15=−a+b−c | |
Let's plug in the second point.
f(1)=a(1)3+b(1)2+c(1) | Simplify from here. |
f(1)=a+b+c | f(1)=−5, so plug that back in! |
−5=a+b+c | |
|
Normally, I would just go right ahead and plug in the next point, but these two equations are somewhat unique.
15=−a+b−c−5=+a+b+c
Look at that! Both the a's and the c's will cancel out here! We can solve for b using elimination.
15=−a+b−c−5=+a+b+c | Add the equations together. |
10=2b | Divide by 2 on both sides! |
5=b | |
Now, let's plug in the third point.
f(x)=ax3+bx2+cx | Evaluate the function at f(2) |
f(2)=a(2)3+b(2)2+c(2) | Now, simplify. |
f(2)=8a+4b+2c | Remember that f(2)=12 and b=5. |
12=8a+20+2c | Subtract 20 from both sides. |
−8=8a+2c | Divide by 2 because that is the GCF of the coefficients, and it will make the numbers nicer in the long run. |
−4=4a+c | |
Ok, let's compare the previous equation with the first one we calculated.
15=−a+b−c−4=4a+c
It looks like more cancellation can be done, right? Well' let's do that then!
15=−a+b−c−4=4a+c | Add both equations together. |
11=3a+b | Of course, b=5. |
11=3a+5 | Now, solve for a. |
6=3a | |
2=a | |
Now, plug this back into an already existing equation to find c.
−4=4a+c | a=2, so substitute that value in. |
−4=4∗2+c | Now, solve for c. |
−4=8+c | Subtract 8 from both sides and isolate c. |
−12=c | |
Ok, our function is now clearly defined for the points given. It is now f(x)=2x3+5x2−12x. It is time to find the remaining zeros!
2x3+5x2−12x=0 | Factor out a GCF first before doing anything else! | |||
x(2x2+5x−12)=0 | The next step involves finding factors of -24 that add up to 5. (-3 and 8). Break up the linear term into these portions. | |||
x(2x2−3x+8x−12)=0 | Now, factor by grouping. | |||
x[x(2x−3)+4(2x−3)] | ||||
x(x+4)(2x−3)=0 | Now, set each factor equal to zero and solve. | |||
| All the zeros have been solved for finally. | |||
Therefore, the zeros are −4,0,1.5.