Pearson Algebra 2 Common Core, 2011
PA
Pearson Algebra 2 Common Core, 2011 View details
End-of-Course Assessment

Exercise 47 Page 968

Plot the points on a coordinate plane. Then, use the vertex form to find the equation of the quadratic function.

F

Practice makes perfect

We are given a table that represents a quadratic function and we want to find an equation that best represents the function.

x f(x)
1 - 13
2 - 3
3 3
4 5
5 3

Let's plot the ordered pairs on a coordinate plane and connect them with a smooth curve.

Since the table represented a quadratic function, we obtained a parabola. Notice that we can identify the coordinates of its vertex.

Because we know the vertex, we can write the equation in a vertex form. f(x)= a(x- h)^2+ k In this expression, a, h, and k are either positive or negative constants. In our case the vertex is ( 4, 5), so we have h= 4 and k= 5. We can use these values to partially write our equation. f(x)= a(x-( 4))^2+( 5) ⇕ f(x)= a(x-4)^2+5 To find the value of a, we will use one of the given points that is not the vertex. Let's take ( 2, - 3). Since this point is on the curve, it satisfies its equation. Therefore, to find the value of a, we can substitute 2 for x and - 3 for f(x) and simplify.
f(x)=a(x-4)^2+5
-3=a( 2-4)^2+5
Solve for a
- 3=a(- 2)^2+5
- 3=a(2)^2+5
- 3=a(4)+5
- 8=a(4)
-8/4=a
- 2=a
a=- 2
We found that a= - 2. Now we can complete the equation of the quadratic function. f(x)= - 2(x-4)^2+5 This corresponds to option F.