Question

Convert from vertex form to standard form. (3 pts). f(x) =(x-3)squared + 6

Answer 1

Answer:

Standard form of $f(x) =(x-3)^2 + 6$ is $\mathbf{x^2-6x+15}$

Step-by-step explanation:

Convert from vertex form to standard form $f(x) =(x-3)^2 + 6$

If the vertex form is: $a(x-h)^2+k$

The Standard form is: $ax^2+bx+c$

Now, converting $f(x) =(x-3)^2 + 6$ into standard form

We know that: $(a-b)^2=a^2-2ab+b^2$

Using the formula:

$f(x) =(x-3)^2 + 6\\f(x)=x^2-2(x)(3)+9+6\\f(x)=x^2-6x+15$

So, Standard form of $f(x) =(x-3)^2 + 6$ is $\mathbf{x^2-6x+15}$