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# 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

- Anonymous8 years agoFavorite 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:

- Tino-sbyLv 58 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

- loyLv 45 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 .