# Solving Problems Involving Quadratic Relations.

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• Nov 13th 2011, 04:23 PM
danman
its on quadratic relations and different forms ie. standard factored vertex exc all word problems (there are 3 questions)
Looking for someone to do this out of generosity
Big Pics

I dont know if these answers are right, feel free to correct me on them.
http://i.imgur.com/55Lk4.jpg
http://i.imgur.com/TOdHi.jpg
• Nov 13th 2011, 05:05 PM
Quacky
Is this something upon which you are being assessed?
• Nov 13th 2011, 05:26 PM
danman
No this is a test review and i need help understand these questions because the test is on monday (tommorow)
• Nov 13th 2011, 05:44 PM
Quacky
Well we have:

\$\displaystyle h=-0.04(d-19)^2+14.4\$
1) Is fine except that I would advise you to label the points of intersection with the axis.
Your factors are correct. The distance is correct.You're missing a minus sign on the \$\displaystyle ax^2+bx+c=0\$ form.

Moving onto 2: The rocket will hit the ground when \$\displaystyle h=0\$

Can you solve
\$\displaystyle -4.9t^2+44.1t=0\$? Take out a common factor of \$\displaystyle t\$ and then solve.

For the max height, why not put the equation into vertex form? Alternatively, it will be the point halfway between the x-intercepts, so you could find this point and then substitute it into the formula.

I don't feel like the COMM part is something I can answer without knowing what exactly the examiners are looking for in an answer - for that, I'd consult a teacher.
• Nov 13th 2011, 06:00 PM
danman
I have fixed number 1, but i am still confused for #2 on how to remove the common factor then solve. If you can further explain 2 than i should be able to turn it into factored form. As for communications mark that does not matter on this, it will matter on the actual test.
• Nov 13th 2011, 06:03 PM
Quacky
\$\displaystyle -4.9t^2+44.1t=0\$

Divide both sides by \$\displaystyle 4.9\$:

\$\displaystyle -t^2+9t=0\$

\$\displaystyle t(9-t)=0\$
• Nov 13th 2011, 06:10 PM
danman
Once i do that do i FOIL so it will equal http://latex.codecogs.com/gif.latex?9t+t%5E2 ?
• Nov 13th 2011, 06:16 PM
Quacky
The equation marks the flight path of a rocket. We already know the equation of the quadratic, which is the height, h(t) relates to the time, t, in seconds after launch. The equation is \$\displaystyle h(t)=-4.9t^2+44.1t\$. The height will be \$\displaystyle 0\$ when \$\displaystyle h(t)=0\$ so you are being asked to solve

\$\displaystyle 0=-4.9t^2+44.1t\$
I noticed a common factor of \$\displaystyle 4.9\$ (knowing my \$\displaystyle 4.9\$ times table has helped immeasurably in life) so divided through both sides by \$\displaystyle 4.9\$. This gave:

\$\displaystyle 0=-t^2+9t\$
There is now a common factor of t, which can be factored:
\$\displaystyle t(-t+9)=0\$
By the null factor law, there are only two ways this can be possible: if \$\displaystyle t=0\$ or if \$\displaystyle -t+9=0 \$
• Nov 13th 2011, 06:23 PM
danman
Ok, but once i get http://latex.codecogs.com/png.latex?t%28-t+9%29=0 What is the next step? because if http://latex.codecogs.com/png.latex?t=0 or http://latex.codecogs.com/png.latex?-t+9=0 than i dont know how to get there.
• Nov 13th 2011, 06:26 PM
Quacky
If you cannot solve:
\$\displaystyle -t+9=0\$, then I am incredibly concerned about your test tomorrow! Can you not see that this means \$\displaystyle t=9\$?
• Nov 13th 2011, 06:33 PM
danman
No i did not see this because i have not done it this exact way before, the teacher does not explain these kinds of things well which is why i have to come online to get the help required.
• Nov 13th 2011, 06:35 PM
Quacky
Ok. What about part b? Can you put the quadratic into vertex form? (Which is the form it was in at the start of q1)
• Nov 13th 2011, 07:00 PM
danman