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Answer :
The equation of the form [tex]f(x)=ax^2+bx+c[/tex] that must solve algebraically and f(1)=4, f(2)=13 and, f(4)=46 is [tex]\frac{5}{2}x^2+\frac{3}{2}x+5[/tex].
Putting x = 1 in the given equation, we get
[tex]f(1)=a(1)^2+b(1)+c[/tex]
f(1) = a + b + c = 4 ...(1)
Putting x = 2 in the given equation, we get
[tex]f(2)=a(2)^2+b(2)+c[/tex]
f(2) = 4a + 2b + c = 13 ...(2)
Putting x = 4 in the given equation, we get
[tex]f(4)=a(4)^2+b(4)+c[/tex]
f(4) = 16a + 4b + c = 46 ...(3)
Using elimination method to solve the set of linear equations, we get
(2) - (1), we get
3a + b = 9 ...(4)
(3) - (2), we get
12a + 2b = 33 ...(5)
Multiplying (4) by 2, we get
6a + 2b = 18 ...(6)
(5) - (6), we get
6a = 15
a = 15/6 = 5/2
Putting a = 5/2 in (4), we get
3(5/2) + b = 9
15/2 + b = 18/2
b = 3/2
Putting the values of a and b in (1), we get
5/2 + 3/2 + c = 9
4 + c = 9
c = 9 - 4
c = 5
Hence, the equation is [tex]\frac{5}{2}x^2+\frac{3}{2}x+5[/tex].
To learn more about elimination method, here:-
https://brainly.com/question/14619835
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