# Tangent to the curve

• Feb 23rd 2009, 03:26 PM
mathamatics112
Tangent 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 PM
TheEmptySet
Quote:

Originally Posted by mathamatics112
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)?

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 PM
JoshHJ
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
\$