Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
2. Section 2.2
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Exercise 166 Page 109

How do you find the vertex when the equation is written in graphing form?

Graphing Form: y=(x-2.5)^2+0.75
Vertex: (2.5,0.75)

Practice makes perfect

Let's complete the square. To do that, we will consider the variable terms on the function's right-hand side. y= x^2-5x+7 To visualize how we complete the square, we draw a generic rectangle where the upper left corner has an area of x^2 and the adjacent rectangles each have an area that is half that of -5x.

Since the upper left corner has an area of x^2, it must be a square with a side length of x. This allows us to factor the adjacent rectangle's areas to - 2.5* x. With this information, we can also determine the area of the lower right rectangle which completes the square.

As we can see, we need to add (-2.5)^2 to complete the square. To keep the equation true, this term must be added to both sides. y+ (-2.5)^2 = x^2+2(-2.5x)+ (-2.5)^2+7 Let's isolate f(x) and simplify.
y+(-2.5)^2=x^2+2(-2.5x)+(-2.5)^2+7
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Solve for y
'y+(-2.5)^2=x^2-2(2.5x)+(-2.5)^2+7
'y+2.5^2=x^2-2(2.5x)+2.5^2+7
y+2.5^2=(x-2.5)^2+7
y+6.25=(x-2.5)^2+7
y=(x-2.5)^2+0.75
When a quadratic function is written in graphing form, we can identify its vertex. Graphing Form:& y=a(x- h)^2+ k Vertex:& ( h, k) Let's identify the vertex in our function. Function:& y=(x- 2.5)^2+ 0.75 Vertex:& ( 2.5, 0.75)