Question

algebra 2
name
© 2015 kuta software llc. all rights reserved.
3.3 - day 3 - convert quadratics from standard form to vertex form
use the information provided to write the vertex form equation of each parabola. identify the vertex.

1. $f(x)=x^2 - 10x + 25$
2. $f(x)=x^2 + 8x + 11$
3. $f(x)=x^2 + 16x + 57$
4. $f(x)=x^2 - 16x + 63$
5. $f(x)=-x^2 + 4x - 8$
6. $f(x)=-x^2 - 12x - 43$

Explanation:

Problem 1: \( f(x) = x^2 - 10x + 25 \)

Step 1: Complete the square

Notice that \( x^2 - 10x + 25 \) is a perfect square trinomial. \( x^2 - 10x + 25=(x - 5)^2 \)

Step 2: Identify the vertex

The vertex form of a parabola is \( f(x)=a(x - h)^2 + k \), where the vertex is \( (h,k) \). For \( f(x)=(x - 5)^2 \), \( h = 5 \) and \( k = 0 \)

Step 1: Complete the square

Take the coefficient of \( x \), which is 8. Half of 8 is 4, and \( 4^2=16 \). We add and subtract 16 inside the function:
\( f(x)=x^2 + 8x+16 - 16 + 11 \)
\( f(x)=(x + 4)^2-5 \)

Step 2: Identify the vertex

For \( f(x)=(x + 4)^2-5 \), using the vertex form \( f(x)=a(x - h)^2 + k \) (here \( h=- 4 \), \( k = - 5 \))

Step 1: Complete the square

The coefficient of \( x \) is 16. Half of 16 is 8, and \( 8^2 = 64 \). Add and subtract 64:
\( f(x)=x^2+16x + 64-64 + 57 \)
\( f(x)=(x + 8)^2-7 \)

Step 2: Identify the vertex

For \( f(x)=(x + 8)^2-7 \), \( h=-8 \), \( k=-7 \)

Answer:

Vertex form: \( f(x)=(x - 5)^2 \), Vertex: \( (5,0) \)

Problem 2: \( f(x)=x^2 + 8x + 11 \)