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anonymous
 5 years ago
how do i find intercepts of a quadratic equation in this form y=(x3)squared  1?
anonymous
 5 years ago
how do i find intercepts of a quadratic equation in this form y=(x3)squared  1?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0set y to zero and solve for the x intercepts set x to zero for the y intercepts

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the intercepts on the graph are 0,3 and 2,3. can i even get that by setting those to 0?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(2,3) is an intercept? that's not possible? unless you meant the intercept with another equation? intercepts include x intercepts and y intercepts (2,3) cannot be a x/y intercept

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah my bad. 2,3 is just the refraction on the otherside of the parabola. but i cant figure out how to get 0,3 from this equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok y=(x3)^2 1 set y = 0 (x  3)^2  1= 0 (x  3)^2 = 1 x  3 = 1 x  3 = 1 x = 4 x = 2 i get the x intercept to be (4, 0) and (2, 0) set x = 0 y = (0  3)^2  1 y = 9  1 y = 8 and the y intercept to be (0, 8) i put it in a graphing calculator and they're right so either I don't understand your question, or you typed it wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0original problem is y1= (x3)^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah it's wrong your equation should have been y = (x3)^2 + 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(0, 3) is not a point on the parabola just try plugging 0 for x it's not possible you get y = (0  3)^2 + 1 y = 9 + 1 y = 10 (0, 10)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now i see. thankyou. one more question. the vertex on a graph given is 1,1...it gives four equations to choose from but the one it says is the answer is h(x)=(x1)^2+1. does only the sign before the k value matter?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no rewrite the equation y = (x1)^2 + 1 and again y  1 = (x1)^2 now i'm not sure if i can make myself clear. the vertex for y = x^2 is (0, 0) yes? it would be the same if we wrote y  0 = (x  0)^2 get it? the vertex for y  1 = (x1)^2 (it shift one unit right and one unit up) would be (1, 1)
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