McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
5. Quadratic Inequalities
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Exercise 68 Page 213

Identify the coordinates of the vertex.

y=- 3(x+2)^2+1

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 ( - 2,1). This means that we have h= - 2 and k=1. We can use these values to partially write our function. y= a(x-( - 2))^2+1 ⇕ y= a(x+2)^2+1 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 (- 3,- 2). Since this point is on the curve, it satisfies its equation. Hence, to find the value of a, we can substitute - 3 for x and - 2 for y and simplify.
y=a(x+2)^2+1
- 2=a( - 3+2)^2+1
â–Ľ
Solve for a
- 2=a(- 1)^2+1
- 2=a(1)+1
- 3=a(1)
- 3=a
a=- 3
We found that a= - 3. Now we can complete the equation of the curve. y= - 3(x+2)^2+1