It's a matter of perspective

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As \(u\) increases, the points \((0,5)\) and \((2,1)\) are moved in such a way that their \(x\)-coordinates stay the same but their \(y\)-coordinates smoothly change to \(0\). They are transformed to \((0,0)\) and \((2,0)\).

The red dashed line is drawn passing through the two points described above. This line becomes horizontal (and lies on the \(x\)-axis) when \(u=1\).

The original parabola is transformed so that it always passes through these two points. It appears to remain as a parabola.

The point \(C\) marked on the parabola remains fixed throughout.

The curve does remain as a parabola as \(u\) increases, so we could say that its “parabolic” shape stays the same. Its orientation remains upward facing.

However, it is transformed so that when \(u=1\) it passes through the points \((0,0)\) and \((2,0)\) rather than the points \((0,5)\) and \((2,1)\). The original parabola does not intersect the \(x\)-axis but the transformed one does.

Each point on the parabola is being transformed in the same way as each point on the straight line.

The equation of the original straight line can be written as \(y=5-2x\). As \(u\) increases, we are gradually subtracting from its \(y\) values until when \(u=1\) the equation of the line has become \(y=0\). At this stage, we have therefore subtracted \((5-2x)\) from the \(y\) value of each point on the line.

Similarly, if we write the equation of the original parabola as \(y=f(x)\) and subtract \((5-2x)\) when \(u=1\), we can say that the transformed parabola has equation \[\begin{equation} y=f(x)-(5-2x).\label{eq1} \end{equation}\]

This type of transformation is called a shear; in this case, it is a shear parallel to the \(y\)-axis. A shear is a particular example of an affine transformation.

We can see that the transformed parabola intersects the \(x\)-axis at \(x=0\) and \(x=2\), so its equation can be written as \[\begin{equation} y=Kx(x-2),\label{eq2} \end{equation}\]

where \(K\) is some constant.

We saw above \(\eqref{eq1}\) that the transformed parabola has equation \(y=f(x)-(5-2x)\), and also \(\eqref{eq2}\) that it can be written as \(y=Kx(x-2)\). Therefore \[f(x)-(5-2x)=Kx(x-2),\] and we can rearrange this to get \[f(x)=5-2x+Kx(x-2).\]

We can find the value of \(K\) by using the third given point on the original parabola \((4,13)\): \[13 = 5-2\times4+4K(4-2)\] and solving this gives \(K = 2\). Thus \[\begin{align*} f(x) &= 5-2x+2x(x-2)\\ &= 2x^2-6x+5 \end{align*}\]

and the equation of the original parabola is \[y = 2x^2-6x+5.\]