What is an equation in slope-intercept form of the line that is perpendicular to the graph of #y=2x+3# and passes through (3, -4)?

The slope of the line perpendicular to the graph of #y=2x+3# is #-1/2#. The perpendicular slope is the negative inverse of the original slope. The product of perpendicular slopes is #-1#, where:

#m_1m_2=-1#,

where:

#m_1# is the original slope #(2)# and #m_2# is the perpendicular slope.

#2m_2=-1#

Divide both sides by #2#.

#m_2=-1/2#

So we now have the slope and we have been given a point #(color(red)3,color(blue)(-4))#.

Find the point-slope form of the perpendicular line.

#y-y_1=m(x-x_1)#

Plug in the known values.

#y-(color(blue)(-4))=-1/2(x-color(red)3)#

#y+4=-1/2(x-3)# #larr# point-slope form.

To convert the point-slope form to slope-intercept form, solve the point-slope form for #y#.

Slope-intercept form is: #y=mx+b#, where #m# is the slope and #b# is the y-intercept.

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

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

Subtract #4# from both sides.j

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

Multiply #4# by #2/2# to get an equivalent fraction with #2# as the denominator.

#y=-1/2x+3/2-4xx2/2#

Simplify.

#y=-1/2x+3/2-8/2#

Simplify.

#y=-1/2x-5/2# #larr# perpendicular slope-intercept form

graph{(y-2x-3)(y+1/2x+5/2)=0 [-11.25, 11.25, -5.625, 5.625]}