If two graphs Y = px2 + q and y = 2x2 − 1 intersect at x =2, find the value of p in terms of q
Correct Option:
B
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 = \(\frac{7 - q}{4}\)
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 = \(\frac{7 - q}{4}\)