for what values of a and b will the parabola y =x^2 + ax + b be tangent to the curve y = x^3 at point (1,1)?

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- Feb 23rd 2009, 03:26 PMmathamatics112Tangent to the curve
for what values of a and b will the parabola y =x^2 + ax + b be tangent to the curve y = x^3 at point (1,1)?

- Feb 23rd 2009, 03:43 PMTheEmptySet

So we are going to need to solve some equations....

I am going to call $\displaystyle f(x)=y=x^3 $

Starting with $\displaystyle y=x^2+ax+b$ we get

$\displaystyle 1=1^2+a(1)+b \implies a=-b$

$\displaystyle f'(x)=3x^2 \implies f'(1)=3$

Now $\displaystyle y'=2x+a$ using the slope above and the x-coordinate we can solve for a

$\displaystyle 3=2(1)+a \implies a =1$ so $\displaystyle b=-1$

$\displaystyle y=x^2+x-1$ - Feb 23rd 2009, 03:46 PMJoshHJ
First you will want to find the derivative of: $\displaystyle f(x) = x^3$. Using the power rule, $\displaystyle f'(x) = 3x^2$ Now plug in the 1 to find the slope of the curve at x = 1. $\displaystyle 3(1)^2 = 3$. Now you will want to find the derivative of the equation $\displaystyle g(x) = x^2 + ax + b$ and set that equal to the derivative of $\displaystyle f(x)$. So $\displaystyle g'(x) = 2x + a$ and $\displaystyle f'(x) = 3x^2.$

$\displaystyle 3x^2 = 2x + a$

Substituting in the 1:

$\displaystyle

\\3 = 2 + a

$

$\displaystyle

\\a = 1

$

Now that you have the a-value, you will want to solve for b. Set $\displaystyle f(1) = g(1)$ to get the b.

$\displaystyle

\\ x^3 = x^2 + x + b

$

$\displaystyle

\\ 1 = 1 + 1 + b

$

$\displaystyle

\\ b = -2

$

Finally, the equation of the parabola tangent to the cubic is:

$\displaystyle

g(x) = x^2 + x - 1

$