Answer :
Given:
The graph of a quadratic function passes through the points (5,31) (3,11) (0,11).
To find:
The equation of the quadratic function.
Solution:
A quadratic function is defined as:
[tex]y=ax^2+bx+c[/tex] ...(i)
It is passes through the point (0,11). So, substitute [tex]x=0,y=11[/tex] in (i).
[tex]11=a(0)^2+b(0)+c[/tex]
[tex]11=c[/tex]
Putting [tex]c=11[/tex] in (i), we get
[tex]y=ax^2+bx+11[/tex] ...(ii)
The quadratic function passes through the point (5,31). So, substitute [tex]x=5,y=31[/tex] in (ii).
[tex]31=a(5)^2+b(5)+11[/tex]
[tex]31-11=a(25)+5b[/tex]
[tex]20=25a+5b[/tex]
Divide both sides by 5.
[tex]4=5a+b[/tex] ...(iii)
The quadratic function passes through the point (3,11). So, substitute [tex]x=3,y=11[/tex] in (ii).
[tex]11=a(3)^2+b(3)+11[/tex]
[tex]11-11=a(9)+3b[/tex]
[tex]0=9a+3b[/tex]
Divide both sides by 3.
[tex]0=3a+b[/tex] ...(iv)
Subtracting (iv) from (iii), we get
[tex]4-0=5a+b-3a-b[/tex]
[tex]4=2a[/tex]
[tex]\dfrac{4}{2}=a[/tex]
[tex]2=a[/tex]
Putting [tex]a=2[/tex] in (iv), we get
[tex]0=3(2)+b[/tex]
[tex]0=6+b[/tex]
[tex]-6=b[/tex]
Putting [tex]a=2,b=-6[/tex] in (ii), we get
[tex]y=(2)x^2+(-6)x+11[/tex]
[tex]y=2x^2-6x+11[/tex]
Therefore, the required quadratic equation is [tex]y=2x^2-6x+11[/tex].