1. ## help plz

Find an equation for the line tangent to the curve at the given point
x^4+y^3=17 at (-3,-4)

Also if you could post the steps it took you to get the answer plz THANKS GUYS!

2. Differentiate implicity the function, then plug $(x,y)=(-3,-4)$, so $y'$ will be the slope of the tangent line.

Apply the point-slope formula and you're done.

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I dunno if this is the procedure, I don't remember these things -.-

3. Originally Posted by flewoveru
Find an equation for the line tangent to the curve at the given point
x^4+y^3=17 at (-3,-4)

Also if you could post the steps it took you to get the answer plz THANKS GUYS!
As Krizalid said:
$x^4+y^3=17$

$4x^3 + 3y^2 \frac{dy}{dx} = 0$

$\frac{dy}{dx} = - \frac{4x^3}{3y^2}$

So at (-3, -4) the derivative is:
$\frac{dy}{dx} = - \frac{4(-3)^3}{3(-4)^2} = \frac{4 \cdot 27}{3 \cdot 16} = \frac{9}{4}$

Now you know the tangent line at (-3, -4) has a slope of 9/4. Thus
$y = mx + b$

$-4 = \frac{9}{4} (-3) + b$

$b = -4 + \frac{27}{4} = \frac{11}{4}$

$y = \frac{9}{4}x + \frac{11}{4}$