If two graphs y = px\(^2\) + q and y = 2x\(^2\) -1 intersect at x = 2, find the value of p in terms q.
The correct answer is B. \(\frac{7-q}{4}\)
Given that the two graphs \(y = px^2 + q\) and \(y = 2x^2 - 1\) intersect at \(x = 2\), we can find the value of \(p\) in terms of \(q\) as follows:
Since the two graphs intersect at \(x = 2\), their \(y\)-values must be equal at this point. So, we can set the two expressions for \(y\) equal to each other and substitute \(x = 2\) to get:
\(px^2 + q = 2x^2 - 1\)
\(p(2)^2 + q = 2(2)^2 - 1\)
\(4p + q = 8 - 1\)
\(4p + q = 7\)
Solving for p in terms of q, we get:
\(p = \frac{7 - q}{4}\)
So, the value of p in terms of q is \(\frac{7 - q}{4}\).
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