But how would it be sketched without the CAS?
Find the turning points firstly:
\\\frac{dy}{dx}&=0\\\Rightarrow x&=0\end{align*})
This is a quartic polynomial with one stationary point.
Now let's find any axes-intercepts:
&=1\end{align*})
So we have a \(y\)-intercept at \(\left(0,1\right)\), which also happens to be the stationary point. Note that there are no \(x\)-intercepts.
Theoretically, there are a few possible shapes of quartics with one stationary point - namely, a curve similar to the standard quartic/quadratic with a stationary point and no points of inflection, or a curve with a stationary point and
one non-stationary point of inflection. If we consider the underlying quadratic, \(y=x^2+1\), we could potentially sketch this as the product of two functions (where both are \(y=x^2+1\)).
However, a much simpler way of identifying the shape is by considering that \(y=x^2+1\) is symmetrical about the \(y\)-axis. This means that when we square the whole thing, it will
still be symmetrical about the \(y\)-axis, thus, the shape is similar to that of \(y=x^4\) (i.e. it can't be a quartic with a stationary point and a non-stationary point of inflection, otherwise it would lose this symmetry property).
To get the curvature/gradient relatively accurate, you could then plot a few points and draw a smooth basic quartic/quadratic-shaped curve through them.
Note that the shape is different to that of \(y=x^4\) in that \(y=x^4+2x^2+1\ne x^4+1\), but the main difference is how steep the curve is._________________________________________________________________________________________
Can't believe I didn't think to mention this initially. This can be sketched reasonably easily using addition of ordinates (i.e. sketch \(y=x^4+1\) and \(y=2x^2\) then add the curves together using the addition of ordinates technique.
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