What is the equation of the tangent line of #f(x)=6x-x^2 # at #x=-1#?

Question

Cameron Alexander · Accepted Answer

#y=8x+1# #"we require the slope m and a point "(x,y)" on the line"# #•color(white)(x)m_(color(red)"tangent")=f'(-1)# #rArrf'(x)=6-2x# #rArrf'(-1)=6+2=8# #"and "f(-1)=-6-1=-7rArr(-1,-7)# #rArry+7=8(x+1)# #rArry=8x+1larrcolor(red)"equation of tangent"#

Christopher Agresta · Answer

See below:

First step is finding the first derivative of #f#. #f(x)=6x-x^2# #f'(x)=6-2x# Hence: #f'(-1)=6+2=8# The value of 8's significance is that this is the gradient of #f# where #x=-1#. This is also the gradient of the tangent line that touches the graph of #f# at that point. So our line function is currently #y=8x# However, we must also find the y-intercept, but to do this, we also need the y coordinate of the point where #x=-1#. Plug #x=-1# into #f#. #f(-1)=-6-(1)=-7# So a point on the tangent line is #(-1,-7)# We can now determine the line's equation by applying the gradient formula: gradient#=(Deltay)/(Deltax)# Hence: #(y-(-7))/(x-(-1))=8# #y+7=8x+8# #y=8x+1#

William Abdalla · Answer

#=>f(x) = 8x+1#

We receive

#f(x) = 6x - x^2#

Finding the tangent line's slope, locating a point on the line, and writing down the tangent line equation are the steps involved in determining the tangent line equation.

We take the derivative of our function to find the tangent line's slope.

#f'(x) = 6 - 2x#

Substituting our point #x = -1#

#f'(-1) = 6 - 2(-1) = 6+2= color(blue)(8)#

Now that we have our slope, we need to find a point on the line. We have an #x#-coordinate, but we need a #f(x)# too.

#f(-1) = 6(-1) - (-1)^2 = -6 - 1 = -7#

So the point on the line is #color(blue){(-1, -7)}#.

We can find the equation of the line given its slope and a point on it.

#y-y_p = m(x-x_p)#

#y - (-7) = 8(x - (-1))#

#y + 7 = 8x + 8#

#y = 8x + 1#

Hence, the tangent line equation is: #color(blue)(f(x) = 8x+1)#

Isaiah Allen · Answer

undefined The equation of the tangent line of f(x)=6x-x^2 at x=-1 is y = -7x - 1.