How do you find the intercepts for #y=x^2-9x+20#?

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Answer 1 · 2016-02-13T18:05:21

You set #y=0# for the #x# intercepts and #x=0# for the #y# intercepts. Then you solve the equations.

#x=5#
#x=4#
#y=20#

y intercepts

#y=0#

#x^2-9x+20=0#

#Δ=b^2-4ac#

#Δ=(-9)^2-4*1*20=1#

#x_(1,2)=(-b+-sqrt(Δ))/(2a)#

#x_(1,2)=(-(-9)+-sqrt(1))/(2*1)#

#x_(1,2)=(9+-1)/2#

#x_1=5# 1st y intercept

#x_2=4# 2nd y intercept

graph{x^2-9x+20 [-6.84, 15.67, -2, 9.27]}

x intercepts

#x=0#

#y=0^2-9*0+20#

#y=20# 1st (and only) x intercept

graph{x^2-9x+20 [-28.64, 29.12, -0.73, 28.17]}