User:Volvagia356/Sandbox

Description of Parabola

$f(x)=ax^{2}+bx+c\,$

$f(x)=ax^{2}+bx+c$

$f(x)=a(x-p)^{2}+q$

$x+p=0$

$x=-p$

${\frac {b\pm {\sqrt {b^{2}-4ac}}}{2a}}$

First Function

$4=a(-2)^{2}+b(-2)+4.5$

$4=4a-2b+4.5$

$4=4a+4.5$

$4a=-{\frac {1}{2}}$

$a=-{\frac {1}{8}}$

$y=-{\frac {1}{8}}x^{2}+4.5$

Second Function

${\frac {1}{2}}$

${\frac {1}{2}}=0a^{2}+0b+c$

$c={\frac {1}{2}}$

$y=-{\frac {1}{8}}x^{2}+{\frac {1}{2}}$

Third Function

$-{\frac {1}{8}}(x-2)^{2}+{\frac {1}{2}}$

Finding the Surface Area

$\int _{-2}^{2}-{\frac {1}{8}}x^{2}+{\frac {1}{2}}\,dx$

$=\int _{-2}^{2}{\frac {1}{2}}\,dx-{\frac {1}{8}}\int _{-2}^{2}x^{2}\,dx$

$=\int _{-2}^{2}{\frac {1}{2}}\,dx-\left[{\frac {x^{3}}{24}}\right]_{-2}^{2}$

$=\left[{\frac {x}{2}}-{\frac {x^{3}}{24}}+c\right]_{-2}^{2}$

integral of $x^{2}$ is ${\frac {x^{3}}{3}}$

integral of 1/2 is ${\frac {x}{2}}$

$=\left({\frac {2}{2}}-{\frac {2^{3}}{24}}+c\right)-\left({\frac {-2}{2}}-{\frac {-2^{3}}{24}}+c\right)$

$={\frac {4}{3}}$

$={\frac {4}{3}}\operatorname {m} ^{2}$

Volume & Cost of Arch

Structure 1 $4\operatorname {m} ^{2}-{\frac {4}{3}}\operatorname {m} ^{2}={\frac {8}{3}}\operatorname {m} ^{2}$

${\frac {8}{3}}\operatorname {m} ^{2}\times 0.4\operatorname {m} ={\frac {16}{15}}\operatorname {m} ^{3}$

${\frac {16}{15}}\operatorname {m} ^{3}\times \operatorname {RM} 840=\operatorname {RM} 896$

Structure 2

$1\operatorname {m} \times 4\operatorname {m} =4\operatorname {m} ^{2}$

${\frac {1}{2}}\times 0.5\operatorname {m} \times 4\operatorname {m} =1\operatorname {m} ^{2}$

$4\operatorname {m} ^{2}-1\operatorname {m} ^{2}=3\operatorname {m} ^{2}$

$3\operatorname {m} ^{2}\times 0.4\operatorname {m} ^{2}={\frac {6}{5}}\operatorname {cm} ^{3}$

${\frac {6}{5}}\operatorname {cm} ^{3}\times \operatorname {RM} 840=\operatorname {RM} 1008$