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In Problems 9–14, (a) graph each quadratic function by determining whether its graph opens up or down and by finding its vertex, axis of symmetry, y-intercept, and x-intercepts, if any. (b) Determine the domain and the range of the function. (c) Determine where the function is increasing and where it is decreasing.

fx=-12x2+2

Short Answer

Expert verified

Part (a) Part (a) The graph is

Part (b). Domain :-,

Range :(-,2]

Part (c). Decreasing :(-,0)

Increasing :(0,)

Step by step solution

01

Step 1. Given information.

The given function is fx=-12x2+2.

02

Part (a) Step 1. Compare the function with standard form.

The standard form of quadratic function isfx=ax2+bx+c;a0.

So, a=-12,b=0,c=2.

Now determine weather the parabola open up or down a<0,ora>0.

Here a<0, so parabola open down.

03

Part (a) Step 2. Determine the vertex.

The vertex is -b2a,f-b2a.

So, -b2a=-02·-12=0

Also f-b2a=f(0)=-12(0)2+2=2

Then, the vertex has coordinates 0,2.

And the axis of symmetry isx=-b2a=0.

04

Part (a) Step 3. Find x-intercept and y-intercept.

First check the discriminant and then find the x-intercept and y-intercept.

Discriminant: D=b2-4ac=(0)2-4·(-12)·2==4>0

So, D>0which means that the graph of the given function has two x-intercept.

To find x-intercept solve f(x)=0or-12x2+2=0.

x1,2=-b±b2-4ac2ax1,2=0±02-4·-12·22-12x1,2=±4-1x1=2,x2=-2

So the x-intercept are x1=2,x2=-2.

And y-interceptf(0)=02+2=2.

05

Part (a) Step 4. Graph the function. 

The graph of the function is

06

Part (b) Step 1. Determine the domain and the range of the function.  

The domain of function is the set of all real numbers. Based on the graph, the range of function is the interval(-,2].

07

Part (c) Step 1. Determine where the function is increasing and where it is decreasing. 

The function f is increasing on the interval-,2and is decreasing on the interval0,.

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