Given:
The graph of
[tex]y=2x^2-kx+6[/tex]Required:
What are the possible value(s) of k?
Explanation:
[tex]Set\text{ y = 0, evaluate the quadratic at }h=-\frac{b}{2a}and\text{ solve for k}[/tex]You want to find the value the value of k such that the y coordinate of the vertex is 0.
[tex]\begin{gathered} y=2x^2-kx+6 \\ 0=2x^2-kx+6 \end{gathered}[/tex]The x coordinate, h , of the vertex is found, using the following equation:
[tex]\begin{gathered} D=b^2-4ac \\ b^2-4ac=0 \\ k^2-4\times2\times6=0 \\ k^2-48=0 \\ k^2=48 \\ k=\pm4\sqrt{3} \end{gathered}[/tex]Answer:
So, values of k are above.
