Hi! Are all of the \(-\left(x+y\right)^2\ \left(x-y\right)^2\) like equations that have only a square of like one of these (x-3) all only give the turning point? (instead of the x-intercepts because both of them give the same x intercept, like \(-\left(x-1\right)^2)\
If an equation gives me the two same x intercepts, does that mean that the turning point must lie on the x-axis?
I think I know what you're trying to ask, but I'll need you to clarify.
Are you essentially asking whether the graph of something of the form \(y=(x-a)^2(x-b)(x-c)\) will always have a turning point at \((a,0)\) and will always cut the \(x\)-axis at \((b,0)\) and \((c,0)\)? If so, the answer is
yes.
To answer your second question:
yes. If a polynomial \(P(x)\) has a double root at say \(x=k\), the graph of \(y=P(x)\) will have a turning point at \((k,0)\).
In fact, we can generalise this for roots of higher
multiplicity.
Suppose that for a polynomial \(P(x)\), \(\alpha\) is a root with
even multiplicity and \(\beta\) is a root with
odd multiplicity of at least 3.
That is, \(P(x)=(x-\alpha)^{2k}(x-\beta)^{2k+1}S(x)\), where \(k\in\mathbb{N}\) and \(S(x)\) is another polynomial.
Then, the graph of \(y=P(x)\) will have a turning point at \((\alpha,0)\), and a stationary point of inflection at \((\beta,0)\).
Clearly, as mentioned, \(y=P(x)\) will cut the \(x\)-axis at single roots.