Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
Mid-Chapter Quiz
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Exercise 20 Page 224

Identify the coordinates of the vertex.

y=- (x+2)^2+4

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 ( - 2, 4). This means that we have h= - 2 and k= 4. We can use these values to partially write our function. y= a(x-( - 2))^2+ 4 ⇕ y= a(x+2)^2+4 We can see in the graph that the parabola opens downwards. Therefore, 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 (- 4,0). Since this point lies on the curve, it satisfies its equation. Hence, to find the value of a, we can substitute - 4 for x and 0 for y and simplify.
y=a(x+2)^2+4
0=a( - 4+2)^2+4
Solve for a
0=a(- 2)^2+4
0=a(4)+4
- 4=a(4)
- 1=a
a=- 1
We found that a= - 1. Now we can complete the equation of the curve. y= - 1(x+2)^2+4 ⇔ y=- (x+2)^2+4