Tangent to the curve

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February 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)?
February 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 $b=-1$

$y=x^2+x-1$
February 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:
$<br /> \\3 = 2 + a<br />$
$<br /> \\a = 1<br />$

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

$<br /> \\ x^3 = x^2 + x + b<br />$
$<br /> \\ 1 = 1 + 1 + b<br />$
$<br /> \\ b = -2<br />$

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

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