McGraw Hill Integrated II, 2012
MH
McGraw Hill Integrated II, 2012 View details
Practice Test
Continue to next subchapter

Exercise 11 Page 161

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 ( 0, 2). This means that we have h= 0 and k= 2. We can use these values to partially write our function. y= a(x-( 0))^2+ 2 ⇕ y= ax^2+2 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 a point that lies on the parabola. Let's choose the point (1,- 1).

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