Applications-Of-Derivatives Question 453
Question: The tangent to the curve $ y=ax^{2}+bx $ at $ (2,,-8) $ is parallel to x-axis. Then
[AMU 1999]
Options:
A) $ a=2,,b=-2 $
B) $ a=2,b=-4 $
C) $ a=2b=-8 $
D) $ a=4,,b=-4 $
Show Answer
Answer:
Correct Answer: C
Solution:
$ y=ax^{2}+bx $ $ \frac{dy}{dx}=2ax+b\Rightarrow {{( \frac{dy}{dx} )}_{(2,,-8)}}=4a+b $ $ \because $ Tangent is parallel to x-axis \ $ \frac{dy}{dx}=0\Rightarrow b=-4a $ ?..(i) Now, point (2, ?8) is on the curve of $ y=ax^{2}+bx $ \ $ -8=4a+2b $ ??(ii) From (i) and (ii), we get $ a=2,b=-8 $ .
BETA
