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anonymous

  • 5 years ago

how do i find intercepts of a quadratic equation in this form y=(x-3)squared - 1?

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  1. anonymous
    • 5 years ago
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    set y to zero and solve for the x intercepts set x to zero for the y intercepts

  2. anonymous
    • 5 years ago
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    the intercepts on the graph are 0,3 and 2,3. can i even get that by setting those to 0?

  3. anonymous
    • 5 years ago
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    (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

  4. anonymous
    • 5 years ago
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    yeah 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

  5. anonymous
    • 5 years ago
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    ok y=(x-3)^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

  6. anonymous
    • 5 years ago
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    original problem is y-1= (x-3)^2

  7. anonymous
    • 5 years ago
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    yeah it's wrong your equation should have been y = (x-3)^2 + 1

  8. anonymous
    • 5 years ago
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    (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)

  9. anonymous
    • 5 years ago
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    now 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)=(x-1)^2+1. does only the sign before the k value matter?

  10. anonymous
    • 5 years ago
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    no rewrite the equation y = (x-1)^2 + 1 and again y - 1 = (x-1)^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 = (x-1)^2 (it shift one unit right and one unit up) would be (1, 1)

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