Quadratic Equations:

y = (ax^2)+bx+c

remember linear equations:

y = mx+b

Plot[{3*(x^2)-2x+1,-2x^2+5x+1,x^2-4x-4},{x,-5,5}, PlotRange->{{-5,5},{-5,5}}]

NSolve[3*(x^2)-2x+1==0,x]

Solve[-2x^2+5x+1==0,x]

N[%]

We will now do problem 6 on 5.5

CanData = {{0,1.0},{1,1.25},{2,2.00},{3,3.25},{4,5.0}}

CanPlot =ListPlot[ {CanData }, PlotMarkers->{"X"}, PlotStyle->{Black}, AxesLabel->{"D","H"}, PlotLabel->"Depth of Water as a function of Distance",PlotRange->{{0,6},{0,6}}, PlotLegends->{"Data"}]

CanBestFitLine = LinearModelFit[CanData,{1,t},t]

CanBestFitQuad = LinearModelFit[CanData,{1,t,t^2},t]

Sophie = Plot[CanBestFitLine[t],{t,0,5}];

Hannah = Plot[CanBestFitQuad[t],{t,0,5}];

Show[Sophie, Hannah, CanPlot]

George = Plot[{CanBestFitLine[t],CanBestFitQuad[t]},{t,0,5}];

Show[ CanPlot,George]

CanData

{CanBestFitQuad[0],CanBestFitQuad[1],CanBestFitQuad[2],CanBestFitQuad[3],CanBestFitQuad[4]}

This is the answer to part (b)
If we plug in a distance D, we get out a height H so we solve H(D) = 4 for D.  Remember H is stored in the variable CanBestFitQuad

Solve[4==CanBestFitQuad[D],D]

The answer to part (c):
we have H as a function of D, i.e. H(D), we want to know when D=5, what is H(5)?

CanBestFitQuad[5]

Problem 5 on 5.5

y[g1_, g2_, L_, E_, x_] = (g2-g1)/(2*L)*x^2+g1*x-g1*L/2

y[1.35,-1.75,5,1040.63,x]

y[1.35,-1.75,5,1040.63,0]