If f(x) is a quadratic polynomial such that graph of y=f(x) touches at (4,0) and intersects the positive y−axis at 4, then which of the following is/are correct?
A
f(2)=1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
f(3)=14
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
f(x)=14x2−2x+4
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
f(x)=12x2−x+52
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is Cf(x)=14x2−2x+4 Given f(x) is a quadratic polynomial that touches the x− axis at x=4. ∴f(x)=0 has equal roots equal to x=4, so f(x)=a(x−4)2
Now, the graph of y=f(x) intersect on the positive y−axis f(0)=4⇒a(−4)2=4⇒a=14
Therefore , the required polynomial is, f(x)=14(x−4)2