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mathcalculus

  • one year ago

help! Given determine the points where (a) the curve intercepts the axis; (b) the tangent line to the curve is horizontal.

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  1. mathcalculus
    • one year ago
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  2. mathcalculus
    • one year ago
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    i believe that the 2nd answer is (5,25) right? @dlxhazedxlb @hihihii

  3. hihihii
    • one year ago
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    yup! 2nd answer is (5,25) good job!

  4. mathcalculus
    • one year ago
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    how about the first?

  5. mathcalculus
    • one year ago
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    im not sure how to do this one

  6. mathcalculus
    • one year ago
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    @hihihii

  7. hihihii
    • one year ago
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    in order to find where the curve intersects the x-axis, you have to set the original function equal to 0. -1x^2+ 10x= 0

  8. hihihii
    • one year ago
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    x^2-10x=0 x(x-10)=0 x=0, x=10 to find the points, plug 0 and 10 back into the original equation

  9. hihihii
    • one year ago
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    0+0=0 (0,0) -1(10)^2-10(10)=0 (10,0) both the y values are 0 because that's where it crosses the x-axis

  10. mathcalculus
    • one year ago
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    @hihihii i see where you got x=0 but how'd you get -10?

  11. mathcalculus
    • one year ago
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    i got -x+10

  12. mathcalculus
    • one year ago
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    and also which is the correct answer to A?

  13. mathcalculus
    • one year ago
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    (0,0) or (10,0)? which one do i pick?

  14. mathcalculus
    • one year ago
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    @hihihii

  15. mathcalculus
    • one year ago
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    x^2-10x=0 ??

  16. mathcalculus
    • one year ago
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    hello hello hello

  17. mathcalculus
    • one year ago
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    @satellite73 @SithsAndGiggles would anyone kindly like to help?

  18. SithsAndGiggles
    • one year ago
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    \[f(x)=-x^2+10x\] Finding where the curve intercepts the x-axis is the same as solving for the roots of the equation: \[-x^2+10x=0\] The tangent line to the curve is horizontal when the slope of the tangent line is 0. Hence solve for x in \(f'(x)=0\): \[f'(x)=-2x+10=0\]

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