What are the intercepts of #y = 4(x - 5)+x^2#?

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Answer 1 · 2017-04-16T21:31:47

The x-intercepts are the point of #(2.899,0)# and #(-6.899,0)#, the y-intercept is #(0,-20)#

For y-intercept(s), let #x=0#

Doing so,

#y=4(0-5)+0^2#

#y=4(-5)#

#y=-20#

Therefore the y-intercept is #(0,-20)#

For x-intercept(s), let #y=0#

Doing so,

#0=4(x-5)+x^2#

#x^2+4x-20=0#

Use the quadratic formula (i'll let you do that),

#x_1=-6.899# and #x_2=2.899#

Therefore the x-intercepts are the point of #(2.899,0)# and #(-6.899,0)#, the y-intercept is #(0,-20)#