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Big Ideas Math Algebra 1 A Bridge to Success

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Big Ideas Math Algebra 1 A Bridge to Success View details

Chapter Test

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Exercise 1 Page 473

Identify the vertex and the axis of symmetry of the parabola. Then make a table to find points on the curve.

Graph:

Comparison: The graph of h is a vertical stretch by a factor of 2 and a vertical translation down 3 units of the graph of f.

Let's start by recalling the vertex form of a quadratic function. f(x)=a(x- h)^2+ k In this form the vertex of the parabola is the point ( h, k), and the axis of symmetry is the vertical line x= h. Now consider the given function.

h(x)=2x^2-3 ⇕ h(x)=2(x- 0)^2 + ( - 3) We can see that h= 0 and that k= - 3. Therefore, the vertex is ( 0, - 3), and the axis of symmetry is x= 0. To graph the function we will make a table of values. Make sure to include x-values to the left and to the right of the axis of symmetry.

x	2x^2-3	h(x)=2x^2-3
- 2	2( - 2)^2-3	5
- 1	2( - 1)^2-3	- 1
1	2( 1)^2-3	- 1
2	2( 2)^2-3	5

Let's now draw the parabola that connects the points and the vertex. We will also draw the axis of symmetry x=0, and the parent function f(x)=x^2.

From the graph above, we can note the following.

Both graphs open up.
Both graphs have the same axis of symmetry x=0.
The graph of the given function is narrower than the graph of the parent function.
The vertex of the given function, (0,- 3), is below the vertex of the parent function, (0,0).

From the graph and the observations above, we can conclude that the graph of h is a vertical stretch by a factor of 2 and a vertical translation down 3 units of the graph of f.

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