please help me solve this question
step by step
We know that the expression for polynomial function f(x) having
roots at a,b,c..... is given by f(x)=k(x-a)(x-b)(x-c)..... where k
is the leading coefficient of the polynomial function. Also, a
quadratic function is a polynomial function and if a polynomial
function has zeros having even multiplicity then at that point the
graph does NOT cross x axis and if it has odd multiplicity, then it
passes x axis.
From the given graph, we observe the following things:
f(x) has 0, 2 as the zeros.
At both 0 and 2, the graph crosses x axis and hence it has zeros
having odd multiplicities. Since we have to find the quadratic
function, the zeros will be 0 and 2.
The graph of quadratic function passes through the point (1,5),
hence the polynomial must satisfy the given point.
Now by using the above observations, we may write the expression
for polynomial function f(x) as:-
f(x)=k(x-0)(x-2)
=> f(x)=k(x)(x-2)
f(x) must satisfy (1,5)
=> 5=k(1)(1-2)
=> k=-5
Hence the polynomial function is:-
=> f(x)=-5(x)(x-2)
Now, we have to represent the above function in the form a(x-h)2+k
=> f(x)=-5(x)(x-2)
=> f(x)=-5(x2-2x)
=> f(x)=-5{(x)2-2(x)(1)+12-12}
=> f(x)=-5{(x-1)2-1}
=> f(x)=-5(x-1)2+5
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