Question 147135
<font size = 8 color="red"><b>Edwin's solution:</b></font>

Find the verex and intercepts. Sketch the graph
{{{g(x) = x^2 -6}}}
<pre><font size = 4 color = "indigo"><b>
The vertex formula:

The graph of 

{{{f(x) = ax^2+bx+c}}}

has the vertex ({{{-b/(2a)}}}, {{{-d/(4a)}}}), where {{{d = discriminant = b^2-4ac}}}

Therefore
{{{g(x) = x^2 -6}}}

can be written as

{{{g(x) = x^2+0x-6}}}

So we apply the vertex formula:

{{{a=1}}}, {{{b=0}}}, {{{c=-6}}}, {{{d=discriminant=b^2-4ac=0^2-4(1)(-6)=24}}}

so the vertex is

({{{-b/(2a)}}}, {{{-d/(4a)}}}) =

({{{-0/(2(1))}}}, {{{-24/(4(1))}}}) = ({{{0)}}}, {{{-6}}})

To find the y-intercept, we substitute {{{x=0}}}

{{{g(x)= x^2 -6}}}

{{{g(0) = 0^2 -6}}}

So the y-intecept is ({{{0}}}, {{{-6}}}), which just happens
to, in this case, be the same point as the vertex.

To find the x-intercepts, we we substitute {{{g(x)=0}}}

{{{g(x)= x^2 -6}}}
{{{0= x^2 -6}}}
{{{-x^2=-6}}}
{{{x^2=6}}}
{{{x}}} = ±{{{sqrt(6)}}}
 
Since the square root of 6 is about {{{2.45}}},

the two x intercepts are about ({{{-2.45}}}, {{{0}}}) and ({{{2.45}}}, {{{0}}})

So we plot those two points, as well as:

vertex: ({{{0)}}}, {{{-6}}})

and

y-intercept: ({{{0)}}}, {{{-6}}}) [Which in this case just 
happens to be the same point as the vertex.]

{{{drawing(400,375,-6,6,-8,4, 

graph(400,375,-6,6,-8,4,x^2-6),

locate(0,-6,"(0,-6)"),

locate(-sqrt(6)-2.5,.7,"(-2.45,0)\"),
locate(sqrt(6)+.3,.7,"/(2.45,0)")


 )}}}

Edwin</pre>