# calculus/derivative question

• Oct 15th 2008, 08:53 PM
sxusteven
calculus/derivative question
for what values of a and b is the line 2x+y=b tangent to the curve at y=ax^2 at x=-4

help would be welcome, thanks in advnace (Nod)

• Oct 15th 2008, 08:56 PM
mr fantastic
Quote:

Originally Posted by sxusteven
for what values of a and b is the line 2x+y=b tangent to the curve at y=ax^2 at x=-4

help would be welcome, thanks in advnace (Nod)

At x = -4 the gradient of the tangent is to the curve is -8a. So your line has to have this gradient ......

Also, at x = -4, y = 16a. So the point (-4, 16a) lies on the curve and hence must also lie on the line .....
• Oct 15th 2008, 08:59 PM
Chris L T521
Quote:

Originally Posted by sxusteven
for what values of a and b is the line 2x+y=b tangent to the curve at y=ax^2 at x=-4

help would be welcome, thanks in advnace (Nod)

Rewrite the equation of the line as \$\displaystyle y=-2x+b\$

The slope of this line is \$\displaystyle -2\$

Now, differentiate the function \$\displaystyle y=ax^2\$to get \$\displaystyle y'=2ax\$

At x=-4, the value of the derivative is \$\displaystyle -8a\$.

Now when is \$\displaystyle -2=-8a\$?

Once you have a, then plug the point \$\displaystyle x=-4\$ into the function \$\displaystyle y=ax^2\$ to find the value of the function at this point [which I will call \$\displaystyle y_0\$].

Then, take this point and plug it into the equation for the line:

\$\displaystyle (y_0)=-2(-4)+b\$, where \$\displaystyle (-4,y_0)\$ is the point on the function (and on the line)

Then solve for b.

Can you take it from here?

--Chris