How do you find the x and y intercept given #f(x) = 2/3x + 4#?

2 Answers

Answer:

x-intercept: #=-6#

y-intercept: #=4#

Explanation:

The given equation of straight line is

#f(x)=2/3x+4#

#y=2/3x+4#

#-2/3x+y=4#

#1/4(-2/3x+y)=1#

#-1/6x+y/4=1#

#x/{-6}+y/4=1#

Above equation is in standard intercept form: #x/a+y/b=1#

which has

x-intercept: #a=-6#

y-intercept: #b=4#

Jul 28, 2018

Answer:

The intercepts are #(-6,0)# on the #x#-axis
and #(0,4)# on the #y#-axis.

Explanation:

#f(x) = 2/3x+4# can be written as # y = 2/3x+4#

This is in slope-intercept form, #y = mx +c# so we immediately know that the #y# -intercept is #4#, the point #(0,4)#
On the #y# -axis the #x# -value is always #0#.

To find the #x#- intercept, set #y=0# because on the #x#-axis the #y#-value is always #0#

#0=2/3x +4" "larr xx 3#

#0 = 2x+12#

#-12 = 2x#

#-6=x" "larr# this is the #x#-intercept

The two intercepts are #(-6,0) and (0,4)#

We could also have changed the original equation to :

#3y = 2x +12" "# which is re-arranged to:

#2x-3y = -12" "larr# this is standard form.

Set #x =0# to get # y = 4" "larr# the #y#-intercept

Set #y=0# to get # x =-6" "larr# the #x#-intercept