Here's the question you clicked on:
abgl20
Determine the points of intersection of two graphs defined by y = x^2 + 1 and y = 3x + 5
there are two points of intersection, (2,-1) and (4,17)...u could have just put it in the calculator!
thanks, but can you kindly show me the solution?
I'm sorry I will should how you how to get the solution...or I will walk you through it :)
So we have y=x^2+1 and y=3x+5 We want to know when these are these same So we want to know when x^2+1 is the same as 3x+5 x^2+1 = 3x+5 Can you solve x^2+1=3x+5
\[x^2+1-3x-5=0\] \[x^2-3x-4=0\] How about this? This form should look familiar
x=-1 x=4 are those right? btw, thank you for the patience, freckles..
(x-4)(x+1)=0 yes thats right! :)