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If two graphs y = px2 + q and y = 2x2 − 1 intersect at x =2, find the value of...


Question

If two graphs y = px2 + q and y = 2x2 − 1 intersect at x =2, find the value of p in terms of q

Options

A) q − \(\frac{8}{7}\)

B) 7 − \(\frac{q}{4}\)

C) 8 − \(\frac{q}{2}\)

D) 7 + \(\frac{q}{8}\)

The correct answer is B.

Explanation:

y = px2 + q
y = 2x2 - 1
px2 + q = 2x2 - 1
px2 = 2x2 - 1 - q
p = \(\frac{2x^2 - 1 - q}{x^2}\)
at x = 2
p = \(\frac{2(2)^2 - 1 - q}{2^2}\)
= \(\frac{2(4) - 1 -q}{4}\)
= \(\frac{8 - 1 - q}{4}\)
p = \(\frac{7 - q}{4}\)

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Dicussion (1)

  • y = px2 + q
    y = 2x2 - 1
    px2 + q = 2x2 - 1
    px2 = 2x2 - 1 - q
    p = \(\frac{2x^2 - 1 - q}{x^2}\)
    at x = 2
    p = \(\frac{2(2)^2 - 1 - q}{2^2}\)
    = \(\frac{2(4) - 1 -q}{4}\)
    = \(\frac{8 - 1 - q}{4}\)
    p = \(\frac{7 - q}{4}\)

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