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Mathematics 15 Online
OpenStudy (anonymous):

WILL MEDAL!! The functions f(x) and g(x) are described using the following equation and table: f(x) = -3(1.02)^x x -1 0 1 2 gx -5 -3 -1 1 Which statement best compares the y-intercepts of f(x) and g(x)? A. The y-intercept of f(x) is equal to the y-intercept of g(x). B. The y-intercept of f(x) is equal to 2 times the y-intercept of g(x). C. The y-intercept of g(x) is equal to 2 times the y-intercept of f(x). D. The y-intercept of g(x) is equal to 2 plus the y-intercept of f(x).

OpenStudy (shiraz14):

You shouldn't be posting Exam questions here for answers. A.

OpenStudy (anonymous):

Wait I think I get it is it because -3 would be the y intercept for the 1st equation and then for the second when its 0 its -3 so -3 would also be the y intercept.. @shiraz14

OpenStudy (shiraz14):

@pflorido : Yes, you're correct. Good work that you managed to work that out.

OpenStudy (anonymous):

Thanks ;)

OpenStudy (anonymous):

Can you help me with another? @shiraz14

OpenStudy (anonymous):

Which graph best represents the equation y = the square root of the quantity x plus 8

OpenStudy (anonymous):

OpenStudy (shiraz14):

@pflorido : Where do you think the x-intercept of this graph lies?

OpenStudy (anonymous):

(-8,0)?

OpenStudy (shiraz14):

If you substitute these values of x & y into the equation, you'll find if it satisfies the equation.

OpenStudy (anonymous):

But everytime I plug in one of the x-values into the options the y values never come out as 0

OpenStudy (anonymous):

@shiraz14

OpenStudy (shiraz14):

@pflorido : I'm assuming that your equation looks like this: y= √(x+8) Is this correct?

OpenStudy (anonymous):

yes.. wait would it be graph a?

OpenStudy (anonymous):

Cuz ik the x-intercept is -8 and I just notice that that is the only graph that starts there

OpenStudy (shiraz14):

When x = -8, we get y = √(8-8) = √0 = 0. So yes, you're correct - the answer should be A.

OpenStudy (anonymous):

Thank you so much! I became your fan just for all the help! :)

OpenStudy (shiraz14):

You're welcome

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