Exercise 1 Page 185

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.

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.