Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and the Yahoo Answers website is now in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

Curve Tangent Question?

Hello,

The line y = 16x - 9 is a tangent to the curve y = 2x^3 + ax^2 + bx - 9 at the point (1, 7).

Find the values of a and b.

Thanks in advance.

-Covert

3 Answers

Relevance
  • Anonymous
    8 years ago
    Favorite Answer

    Hello Covert,

    I have posted a full solution to this problem at one of the math forums I help to moderate so that I may give you an easy to read explanation using LaTeX and a diagram to show the result:

    http://www.mathhelpboards.com/f10/coverts-question...

  • 8 years ago

    y = 2x^3 + ax^2 + bx - 9

    This curve must pass point 1,7

    So:

    7 = 2. 1^3 + a. 1^2 + b .1 -9

    7 = 2 + a + b - 9

    14 = a + b .............(1st equation)

    line y = 16x - 9 have gradient m1 = 16

    the curve gradient at point 1,7 is m2 = -1/m1 = -1/16

    curve gradient: y' = 6x^2 + 2ax + b

    -1/16 = 6 + 2a + b

    -1 = 96 + 32a + 16b

    -97 = 32a + 16b ..........(2nd equation)

    With the 1st and 2nd equation, you can get:

    a = -321/16

    b = 545/16

  • loy
    Lv 4
    5 years ago

    y' = 3x^2 + a million = 4 x^2 = a million then x = ±a million tangent line : y = m(x - x0) + y0 y = 4(x-a million) + 2 or y = 4(x+a million) - 2 the smallest slope of the curve is while y' is smallest ... it is while y' = a million considering you have a quadratic first term ... its minimum fee of the 1st term is 0 . this occurs while x = 0 .

Still have questions? Get your answers by asking now.