Answer:
[tex]f(x)=-\frac{4}{7}(x-\sqrt{2}-3)(x+\sqrt{2}-3)[/tex]
or
[tex]f(x)=-\frac{4}{7}(x^2-6x+7)[/tex]
Step-by-step explanation:
we have
f(3+√2 )=f(3− √2 )=0
so
x1=3+√2 and x2=3−√2 are the roots or x-intercepts of the quadratic equation
The quadratic equation in factored form is equal to
[tex]f(x)=a(x-x1)(x-x2)[/tex]
substitute
[tex]f(x)=a(x-(3+\sqrt{2}))(x-(3-\sqrt{2}))[/tex]
Find the value of the coefficient a
Remember that
f(1)=−8
so
For x=-1 -----> f(x)=-8
substitute
[tex]-8=a(-1-(3+\sqrt{2}))(-1-(3-\sqrt{2}))[/tex]
solve for a
[tex]-8=a(-4-\sqrt{2}))(-4+\sqrt{2}))[/tex]
apply difference of squares
[tex]-8=a(14)[/tex]
[tex]a=-\frac{4}{7}[/tex]
therefore
[tex]f(x)=-\frac{4}{7}(x-(3+\sqrt{2}))(x-(3-\sqrt{2}))[/tex]
[tex]f(x)=-\frac{4}{7}(x-\sqrt{2}-3)(x+\sqrt{2}-3)[/tex]
simplify
Applying distributive property
[tex]f(x)=-\frac{4}{7}(x^2+\sqrt{2}x-3x-\sqrt{2}x-2+3\sqrt{2}-3x-3\sqrt{2}+9)[/tex]
[tex]f(x)=-\frac{4}{7}(x^2-6x+7)[/tex]
