McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
7. Transformations of Quadratic Graphs
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Exercise 4 Page 278

Identify the coordinates of the vertex.

A

Practice makes perfect

We want to write the equation of the given parabola. To do so, let's recall the vertex form of a quadratic function. y= a(x- h)^2+k In this expression, a, h, and k are either positive or negative constants. Let's start by identifying the vertex.

The vertex of this parabola has coordinates ( -3,6). This means that we have h= -3 and k=6. We can use these values to partially write our function. y= a(x-( -3))^2+6 ⇕ y= a(x+3)^2+6 We can see in the graph that the parabola opens downwards. Thus, a will be a negative number. To find its value, we will use the given point that is not the vertex.

We can see above that the point has coordinates (-5,2). Since this point is on the curve, it satisfies its equation. Hence, to find the value of a, we can substitute -5 for x and 2 for y and simplify.
y=a(x+1)^2-3
2=a(( -5)+3)^2+6
â–Ľ
Solve for a
2=a(-2)^2+6
2=a(4)+6
-4=a(4)
-1=a
a=-1
We found that a= -1. Now we can complete the equation of the curve. y= -1(x+3)^2+6 The equation above corresponds to the option A.