# Tangent to the curve

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• 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 $f(x)=y=x^3$

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

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

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

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

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

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

$
\\a = 1
$

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

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

$
\\ 1 = 1 + 1 + b
$

$
\\ b = -2
$

Finally, the equation of the parabola tangent to the cubic is:
$
g(x) = x^2 + x - 1
$