+2442 The equation \(h=-16t^2+100t+5\) 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).
| \(20=-16r^2+100t+5\) | Subtract 20 from both sides. |
| \(-16t^2+100t-15=0\) | Use the quadratic formula to find the solutions of this equation. |
| \(t=\frac{-100\pm\sqrt{100^2-4(-16)(-15)}}{2(-16)}\) | Now, simplify from here. |
| \(t=\frac{-100\pm\sqrt{9040}}{-32}\) | 16 is the largest perfect square factor of 9040, so we can simplify the radical further. |
| \(\sqrt{9040}=\sqrt{16*565}=\sqrt{16}\sqrt{565}=4\sqrt{565}\) | |
| \(t=\frac{-100\pm4\sqrt{565}}{-32}\) | -100,4, and -32 all have a GCF of 4, so this answer can still be simplified further. |
| \(t=\frac{-25\pm\sqrt{565}}{-8}\) | |
| \(t_1=6.096\text{sec}\quad t_2=0.154\) | 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. |
+2442 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 \(f(x)=ax^3+bx^2+cx+d\). We can actually solve for d straightaway by plugging in \(f(0)\).
| \(f(x)=ax^3+bx^2+cx+d\) | Plug in \(f(0)\) |
| \(f(0)=a*0^3+b*0^2+c*0+d\) | 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 \(f(x)=ax^3+bx^2+cx\). It only gets harder from here, though. Let's plug in the remaining points, and let's see what happens!
| \(f(x)=ax^3+bx^2+cx\) | 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! |
| \(-5=a+b+c\) | |
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Normally, I would just go right ahead and plug in the next point, but these two equations are somewhat unique.
\(\hspace{2mm}15=-a+b-c\\ -5=+a+b+c\)
Look at that! Both the a's and the c's will cancel out here! We can solve for b using elimination.
| \(\hspace{2mm}15=-a+b-c\\ -5=+a+b+c\) | Add the equations together. |
| \(10=2b\) | Divide by 2 on both sides! |
| \(5=b\) | |
Now, let's plug in the third point.
| \(f(x)=ax^3+bx^2+cx\) | 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.
\(\hspace{1mm}15=-a+b-c\\ -4=4a\hspace{7mm}+c\)
It looks like more cancellation can be done, right? Well' let's do that then!
| \(\hspace{1mm}15=-a+b-c\\ -4=4a\hspace{7mm}+c\) | 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 \(f(x)=2x^3+5x^2-12x\). It is time to find the remaining zeros!
| \(2x^3+5x^2-12x=0\) | Factor out a GCF first before doing anything else! | |||
| \(x(2x^2+5x-12)=0\) | The next step involves finding factors of -24 that add up to 5. (-3 and 8). Break up the linear term into these portions. | |||
| \(x(2x^2-3x+8x-12)=0\) | 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 \(-4,0,1.5\).
