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31 votes
The graph of y= 2x^2 - kx + 6 touches the x-axis. What are the possible value(s) of k?

User Greg Alexander
by
2.8k points

1 Answer

11 votes
11 votes

Given:

The graph of


y=2x^2-kx+6

Required:

What are the possible value(s) of k?

Step-by-step explanation:


Set\text{ y = 0, evaluate the quadratic at }h=-(b)/(2a)and\text{ solve for k}

You want to find the value the value of k such that the y coordinate of the vertex is 0.


\begin{gathered} y=2x^2-kx+6 \\ 0=2x^2-kx+6 \end{gathered}

The x coordinate, h , of the vertex is found, using the following equation:


\begin{gathered} D=b^2-4ac \\ b^2-4ac=0 \\ k^2-4*2*6=0 \\ k^2-48=0 \\ k^2=48 \\ k=\pm4√(3) \end{gathered}

Answer:

So, values of k are above.

User Patrick Lee Scott
by
3.0k points
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