What are the intercepts for #y=x^2+8x+1#?

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Answer 1 · 2018-07-05T23:36:07

#"Intercept" = 1#

#y = x^2 + 8x + 1#

#y = (x + 8)x + 1#

Recall;

#y = mx + c#

Where;

#c = "intercept"#

Comparing both equations;

#c = 1#

Therefore the intercept is #1#