Question

Form quadratic functions given the following information:

Answer 1

Answer:

$f(x)=-\frac{4}{7}(x-\sqrt{2}-3)(x+\sqrt{2}-3)$

or

$f(x)=-\frac{4}{7}(x^2-6x+7)$

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

$f(x)=a(x-x1)(x-x2)$

substitute

$f(x)=a(x-(3+\sqrt{2}))(x-(3-\sqrt{2}))$

Find the value of the coefficient a

Remember that

f(1)=−8

so

For x=-1 -----> f(x)=-8

substitute

$-8=a(-1-(3+\sqrt{2}))(-1-(3-\sqrt{2}))$

solve for a

$-8=a(-4-\sqrt{2}))(-4+\sqrt{2}))$

apply difference of squares

$-8=a(14)$

$a=-\frac{4}{7}$

therefore

$f(x)=-\frac{4}{7}(x-(3+\sqrt{2}))(x-(3-\sqrt{2}))$

$f(x)=-\frac{4}{7}(x-\sqrt{2}-3)(x+\sqrt{2}-3)$

simplify

Applying distributive property

$f(x)=-\frac{4}{7}(x^2+\sqrt{2}x-3x-\sqrt{2}x-2+3\sqrt{2}-3x-3\sqrt{2}+9)$

$f(x)=-\frac{4}{7}(x^2-6x+7)$