How do you find the slope and intercept of #y = -5x + 2#?

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Answer 1 · 2018-08-02T19:02:54

Slope: #-5#

#x#-intercept: #(2/5, 0)#

#y#-intercept: #(0, 2)#

#y = -5x + 2#

This equation is in slope-intercept form:

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Based on the image, we know that the slope is the value multiplied by #x#, so the slope is #-5#.

To find the #x#-intercept, plug in #0# for #y# and solve for #x#:
#0 = -5x + 2#

Subtract #color(blue)2# from both sides:
#0 quadcolor(blue)(-quad2) = -5x + 2 quadcolor(blue)(-quad2)#

#-2 = -5x#

Divide both sides by #color(blue)(-5)#:
#(-2)/color(blue)(-5) = (-5x)/color(blue)(-5)#

#2/5 = x#

#x = 2/5#

The #x#-intercept is at #(2/5, 0)#.

To find the #y#-intercept, plug in #0# for #x# and solve for #y#:
#y = -5(0) + 2#

#y = 0 + 2#

#y = 2#

The #y#-intercept is at #(0, 2)#.

Hope this helps!

Answer 2 · 2018-08-02T21:37:04

Slope #-5#, #x#-int #2/5#, #y#-int #2#

The good thing is that this equation is in slope-intercept form

#y=mx+b#, with slope #m# and a #y#-intercept of #b#. With this in mind, we see that our slope is #-5# and our #y#-intercept is #2#.

What about the #x#-intercept?

The #x#-intercept is the value when #y=0#. We can plug this into our equation to get

#0=-5x+2#

#-5x=-2=>x=2/5#

Therefore, our slope is #-5#, our #x#-intercept is #2/5# and our #y#-intercept is #2#.

Hope this helps!