#F(x) = x^2+6x+8#?

1 Answer
Oct 19, 2017

Answer:

Note that this is not a question; it is just a definition.

Explanation:

Some possible questions:

1. What are the #x# and #y# intercepts for this equation?
a) The #y# intercept is the value of the equation when #x=0#
In this case if #x=0#
#color(white)("XXX")F(0)=0^2+6xx0+8=8#
So the #y# intercept is #8#

b) The #x# intercepts are the values of #x# for which the equation is equal to #0#
In this case if #x^2+6x+8=0#
#color(white)("XXX")#We can factor the expression on the left:
#color(white)("XXX")(x+2)(x+4)=0#
#color(white)("XXX")rarr{:((x+2)=0,color(white)("xx")"or"color(white)("xx"),(x+4)=0), (rarr x=-2,,rarr x=-4) :}#
So the #x# intercepts are #(-2)# and #(-4)#

2. What is the vertex of this equation?
We can convert this equation into vertex form: #F(x)=(x-color(red)a)^2+color(blue)b# with vertex at #(color(red)a,color(blue)b)#
#color(white)("XXX")F(x)#
#color(white)("XXXXX")=x^2+6x+8#

#color(white)("XXXXX")=x^2+6x+9-1#

#color(white)("XXXXX")=(x+3)^2-1#

#color(white)("XXXXX")=(x-color(red)((-3)))^2+color(blue)((-1))#
which is the vertex form with vertex at #(color(red)((-3)),color(blue)((-1)))#

3. Draw the graph of this equation
graph{x^2+6x+8 [-8.836, 2.266, -1.523, 4.024]}