Find the quadratic approximation of resistance versus temperature for the data given in the previous example between $60\backslash$deg F and $90\backslash$deg F$.$ Solution Again, since $75\backslash$deg F is the midpoint, we will use this for To and therefore R$($To$)=110\mathrm{.}2$ To find a$,$ and a$$ two equations can be setup using the endpoints of the data, namely, R$(60\backslash$deg F$)$ and R$(90\backslash$deg F$).106\backslash$deg $+22=110\mathrm{.}2[1+\backslash$alpha $,(60-75)+\backslash$alpha $(60-75)211112\mathrm{.}21060110\mathrm{.}2[1+\backslash$alpha $,(90-75)+\backslash$alpha $(90-75)22$ adding these two equations e liminates a$,$ so that we can solve for a$$ a$=-44\backslash$times $10-6(\backslash$deg F$)2$ This can be used in either equation to find the value of a $=0\mathrm{.}00875/\backslash$deg F$.$ Thus, the quadratic approximation for the resistance versus temperature is R$($T$)=110\mathrm{.}2[1+0\mathrm{.}001875($T$-75)-44\mathrm{.}36\backslash$times $10-6($T$-75)^2]$