SOLUTION: please find the equation of the tangent to the curve given by y= x^3 - 5x + 15 at the point where x= -1 I got the answer to be y= -2x +17 but im not sure if this is right. co

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: please find the equation of the tangent to the curve given by y= x^3 - 5x + 15 at the point where x= -1 I got the answer to be y= -2x +17 but im not sure if this is right. co      Log On


   

Question 330494: please find the equation of the tangent to the curve given by
y= x^3 - 5x + 15
at the point where x= -1
I got the answer to be
y= -2x +17
but im not sure if this is right.
could you show working if im incorrect please.thanks
Found 2 solutions by Jk22, Fombitz:
Answer by Jk22(389) About Me  (Show Source):
You can put this solution on YOUR website!
Q : tangent to the curve given by
y= x^3 - 5x + 15
at the point where x= -1 ?

y(-1)=-1+5+15=19
y'(x)=3*x^2-5, y'(-1)=3-5=-2
let ytan(x)=ax+b
ytan(-1)=-a+b=19
ytan'(-1)=a=-2 => b=19+a=17
hence ytan(x)=-2x+17

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
The slope of the tangent line equals the value of the derivative at x=-1.
y=x%5E3-5x%2B15
dy%2Fdx=3x%5E2-5
At x=-1
y=%28-1%29%5E3-5%28-1%29%2B15=-1%2B5%2B15=-19
dy%2Fdx=3%28-1%29%5E2-5=-2
Use the point-slope form of a line, y-yp=m%28x-xp%29
y-%28-19%29=-2%28x-%28-1%29%29
y%2B19=-2%28x%2B1%29
y=-2x-2-19
y=-2x%2B17
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That's what I get too.
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