wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If f(x) is a quadratic polynomial such that graph of y=f(x) touches at (4,0) and intersects the positive yaxis 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)=14x22x+4
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
f(x)=12x2x+52
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C f(x)=14x22x+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(x4)2
Now, the graph of y=f(x) intersect on the positive yaxis
f(0)=4a(4)2=4a=14

Therefore , the required polynomial is,
f(x)=14(x4)2

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Quadratic Polynomials
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon