Question

What's the standard form, vertex, and factored​

Answer 1

Answer:

1) $f(x)=(x+1)(x-3)$

2) $f(x)=(x-1)^2-4$

3) $f(x)=x^2-2x-3$

Step-by-step explanation:

So we have a graph and we know that its roots are at x=-1 and x=3.

We also know the vertex is at (1,-4). With that, we can figure out the three forms.

Factored Form:

The factored form, as given, is:

$f(x)=a(x-r_1)(x-r_2)$

We already know the roots of -1 and 3. So, substitute:

$f(x)=a(x-(-1))(x-3)$

Simplify:

$f(x)=a(x+1)(x-3)$

Now, we just need to figure out a. To do so, we can use the vertex. Since the vertex is at (1,-4), this means that f(1) is -4. So, substitute 1 for x and substitute -4 for f(x):

$-4=a(1+1)(1-3)$

Add and subtract:

$-4=a(2)(-2)$

Multiply:

$-4=-4a$

Divide both sides by -4:

$a=1$

So, our factored form is:

$f(x)=(x+1)(x-3)$

Vertex Form:

The vertex form is:

$f(x)=a(x-h)^2+k$

Where (h,k) is the vertex.

We already know the vertex is (1,-4), so substitute 1 for h and -4 for k.

We also previously determined that a is 1, so substitute that also. So:

$f(x)=(1)(x-(1))^2+(-4)$

Simplify:

$f(x)=(x-1)^2-4$

Standard Form:

To acquire the standard form, simply expand the factored or vertex form. I'm going to expand the factored form:

$f(x)=(x+1)(x-3)$

FOIL:

$f(x)=x^2-3x+x-3$

Combine like terms:

$f(x)=x^2-2x-3$

And we're done!