Question

please help!!!!

Let f(x)=x^2+10x+37 .

What is the vertex form of f(x)?

What is the minimum value of f(x)?


Enter your answers in the boxes.

Vertex form: f(x)= _____

Minimum value of f(x): _____

Answer 1

We can go from standard from to vertex form using completing the square.

Standard form: $\sf ax^2+bx+c$
Vertex form: $\sf a(x-h)^2+k$, where (h, k) is the vertex

Group the first two terms:

$\sf x^2+10x+37$

$\sf (x^2+10x)+37$

Now complete the square inside the parenthesis. Take half of the second term and square it, then add it and subtract it inside the parenthesis(so we don't change the function):

Half of 10 is 5, 5 squared is 25:

$\sf (x^2+10x+25-25)+37$

Factor the perfect square($\sf x^2 + 10x + 25$):

$\sf (x+5)^2-25+37$

Simplify:

$\boxed{\sf (x+5)^2+12}$

This is the function in vertex form.

The minimum value is the vertex, we can find this just by looking at our function in vertex form.

Vertex form: $\sf a(x-h)^2+k$, where (h, k) is the vertex

$\boxed{\sf (x+5)^2+12}$

'h' is -5 and 'k' is 12

So our minimum is:

$\boxed{\sf (-5,12)}$

Answer 2

Thank you for the explanation!

To all future Brainly users: Here is simply the answers so you aren't confused.

Vertex form: f(x) = (x+5)²+12

Minimum value of f(x): 12

I hope that clears up any confusion...

Good luck! :)