Respuesta :

Answer:

Set y = 0, evaluate the quadratic at h=−b2a and solve for k.

k=9 I think..... I'm not completely sure but I think that's how it is

Answer:

[tex]\displaystyle \text{Vertex} = \left(k, 3-k^2\right)[/tex]

Step-by-step explanation:

We are given the quadratic equation:

[tex]y=x^2-2kx+3[/tex]

Where k is a non-zero constant.

And we want to determine the vertex of the parabola in terms of k.

The vertex of a parabola is given by the formulas:

[tex]\displaystyle \text{Vertex}=\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)[/tex]

In this case, a = 1, b = -2k, and c = 3.

Find the x-coorinate of the vertex:

[tex]\displaystyle x=-\frac{(-2k)}{2(1)}=\frac{2k}{2}=k[/tex]

To find the y-coordinate, we substitute the value we acquired back into the equation. So:

[tex]\displaystyle \begin{aligned} y(k)&=(k)^2-2k(k)+3\\&=k^2-2k^2+3\\&=3-k^2\end{aligned}[/tex]

Therefore, our vertex in terms of k is:

[tex]\displaystyle \text{Vertex} = \left(k,3-k^2\right)[/tex]

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