How do you find the x and y intercepts for #f(x) = -x^2 + 6x + 8#?

1 Answer
Apr 26, 2017

Answer:

#x_("intercepts")-> -2 and -4#

#y_("intercept") =+8#

Explanation:

Given:#" "f(x)=-x^2+6x+8#

The use of the term #f(x)# is that of giving a label (name) to a particular mathematical construct. The #f# is the name and the #(x)# bit means it is applied to #x#

So #f(a)# would mean: #-a^2+6a+8#
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#color(blue)("Determine x-intercepts")#

Observe that #2xx4=8" and that "2+4=6#

so we can write: #y=(x+2)(x+4)#

The x-intercept is at #y=0# so :

Set #y=0=(x+2)(x+4)#

Thus for this condition #x=-2 and x=-4#
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#color(blue)("Determine y-intercept")#

When the equation is in form #y=ax^2+bx+c# then

#y_("intercept")=c=+8 color(red)(larr" Shortcut")#

You obtain this value by substituting #0" for "x#