Need help with understanding the formula

[Ultimate_ondori]

Today in class my teacher solved the problem below by converting the quadratic function table into a standard form function, the problem is she does not explain it very well...
I understand how she got the h, k, x, and f(x) from the table but I got lost when when she started to plugin numbers into a formula in which I do not know. It would be very helpful if someone explained how to use h, k, x, and f(x) in order to find the equation of the function in standard form step by step

 

[Dr.Peterson]

Today in class my teacher solved the problem below by converting the quadratic function table into a standard form function, the problem is she does not explain it very well...
I understand how she got the h, k, x, and f(x) from the table but I got lost when when she started to plugin numbers into a formula in which I do not know. It would be very helpful if someone explained how to use h, k, x, and f(x) in order to find the equation of the function in standard form step by step
View attachment 34125

Were there words along with the image?

What would help us most is for you to explain the work as far as you understand it, so we can be sure where you are stuck, and what help you need.

Are you saying you don't understand the vertex form, f(x) = a(x-h)^2 + k? Or is it somewhere later? The basic procedure is to put h and k from the table into that equation, put another point into the resulting equation and solve for a, and then rewrite from vertex form to standard form.

 

[Harry_the_cat]

Today in class my teacher solved the problem below by converting the quadratic function table into a standard form function, the problem is she does not explain it very well...
I understand how she got the h, k, x, and f(x) from the table but I got lost when when she started to plugin numbers into a formula in which I do not know. It would be very helpful if someone explained how to use h, k, x, and f(x) in order to find the equation of the function in standard form step by step
View attachment 34125

The vertex form of the equation is \(\displaystyle f(x)=a(x-h)^2+k\) where h and k are the x- and y-coordinates of the vertex.

So, if h=-1 and k=4 are plugged into the equation that gives:

\(\displaystyle f(x)=a(x-^-1)^2 +4 = a(x+1)^2+4\) (*)

At this stage "a" is still unknown, so we need another point to find "a".

(0, 5) is another point, so if we now put x=0 and f(x)=5 into the equation from (*) we get:

\(\displaystyle 5= a(0+1)^2 +4\) which enables us to work out that \(\displaystyle a=1\)

{Note: You could also have used the point (-3, 8) or (-2, 5) to find "a" but using (0, 5) makes the arithmetic easier because of the 0. Try one of these other points for yourself to see that you'll get the same value for "a".)

So, our equation (in vertex form) is:
\(\displaystyle f(x) = 1(x+1)^2+4\)

Expanding this out gives (in standard form):
\(\displaystyle f(x)= x^2 +2x+5\)

Please respond with any step you do not understand and I can explain it further.