Question

The function below is written in vertex form or intercept form. Rewrite them in standard form and show your work.

y = 5(x+3)^2-4

Answer 1

Answer:

The standard form as $y=5x^2+30x+41$

Step-by-step explanation:

Given: A function which is written in vertex form or intercept form.

We have to re-write it  in standard form that in terms of

Given $y = 5(x+3)^2-4$

Squaring  using $(a+b)^2=a^2+b^2+2ab$ , we get,  

$y=5(x^2+9+6x)-4$

Multiply 5 inside , we get,

$y=5x^2+45+30x-4$

Solving further , we get,

$y=5x^2+30x+41$

Thus , we have obtained the standard form as $y=5x^2+30x+41$

Answer 2

Answer: The standard form of equation will be

$f(x)=5x^2+30x+41$

Step-by-step explanation:

Since we have given that

The vertex form of equation is given by

$y=5(x+3)^2-4$

We need to find the standard form :

Standard form is written as :

$f(x)=ax^2+bx+c$

So, our equation becomes,

$y =5(x+3)^2-4\\\\y=5(x^2+9+6x)-4\\\\y=5x^2+45+30x-4\\\\y=5x^2+30x+41$

Hence, the standard form of equation will be

$f(x)=5x^2+30x+41$