SOLUTION: Determine any axis intercepts; y = "sqrt X" + 8 I'm really not getting this!...I know you set a or y to 0 to solve the other, but I am not understanding these questions =*(

Algebra ->  Linear-equations -> SOLUTION: Determine any axis intercepts; y = "sqrt X" + 8 I'm really not getting this!...I know you set a or y to 0 to solve the other, but I am not understanding these questions =*(      Log On


   

Question 782243: Determine any axis intercepts; y = "sqrt X" + 8
I'm really not getting this!...I know you set a or y to 0 to solve the other, but I am not understanding these questions =*(
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
To find the y-intercept, plug in x = 0 and solve for y

y+=+sqrt%28x%29+%2B+8

y+=+sqrt%280%29+%2B+8

y+=+0+%2B+8

y+=+8

The y-intercept is the point (0,8)

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To find the x-intercept, plug in y = 0 and solve for x

y+=+sqrt%28x%29+%2B+8

0+=+sqrt%28x%29+%2B+8

-8+=+sqrt%28x%29

%28-8%29%5E2+=+%28sqrt%28x%29%29%5E2

64+=+x

x+=+64

BUT, there's a problem. Plugging in x = 64 into y+=+sqrt%28x%29+%2B+8 gives you

y+=+sqrt%28x%29+%2B+8

y+=+sqrt%2864%29+%2B+8

y+=+8+%2B+8

y+=+16

which is NOT y = 0, so clearly x = 64 is an extraneous solution.

So there are NO x-intercepts. This means that the graph does NOT cross the x-axis at all.