Question

Write a quadratic function in standard form whose graph satisfies the given conditions. How would you do 17 if the vertex is (-5,4) Do you use vertex form?

Answer 1

We have the following quadratic in the vertex form

`y=a(x+5)^2+4`

We want to expand the quadratic term, which is

`(x+5)^2`

That's the same thing as

`(x+5)(x+5)`

Then let's do the distributive here

`(x+5)(x+5)=x^2+10x+25`

Now we return to our vertex form, and substitute the quadratic term by the factored out

`y=a(x^2+10x+25)+4`

Be careful, don't forget to multiply all by a

`y=ax^2+10ax+25a+4`

That's the quadratic in the standard form. Which is the value of "a"? We don't know, we need extra information to find out the value of a, if we had one more point we could find it, but using only the vertex the value of "a" can be any number, except for 0.

Let's plot some parabolas: