IB-Function on Bridge(real world application)

[Leah5467]

My first attempt is:-8/9(x-3)(x+3),but the answer should be -8/9(x^2)+8

Why is that? I don't really understand. Please help! Thank you!

 

[ksdhart2]

The two answers are equivalent. Try expanding out your answer and you should see why:

\(\displaystyle -\frac{8}{9} (x-3)(x+3) = -\frac{8}{9}\left( x^2 - 9 \right) = \text{???}\)

 

[Leah5467]

Oh! I see why now! Thank you!

 

[MarkFL]

Using the given coordinate axes, we know the vertex is at (0,8), and the parabola opens down, so we may state:

[MATH]f(x)=-k(x-0)^2+8=-kx^2+8[/MATH] where \(0<k\)

Then knowing the point (3,0) is on the curve, we may write:

[MATH]0=-k(3)^2+8\implies k=\frac{8}{9}[/MATH]
Thus:

[MATH]f(x)=-\frac{8}{9}x^2+8[/MATH]
In order for the truck to fit, we require:

[MATH]f(2)>5[/MATH]
Will it fit?

 

[Leah5467]

Thank you for helping! I think it won't fit,as the truck is f(4)=5,so f(2)<5.

 

[Harry_the_cat]

Thank you for helping! I think it won't fit,as the truck is f(4)=5,so f(2)<5.

Not sure what you mean by \(\displaystyle f(4)=5\)?? \(\displaystyle f(4)\neq5\) and \(\displaystyle f(4)\) is irrelevant here anyway.

\(\displaystyle f(2)\approx 4.44 < 5 \) is the correct reason why the truck won't fit.

 

[Leah5467]

oh i see! Thank you! I forgot how i came up with that ?