The derivative finds the slope of the tangent line. The normal line is perpendicular to the tangent line. EX #3: f ( x ) =5 x 2 Find equation of the tangent line and normal line at x =3. 2.5 Derivative as Tangent Line Notes 2.5 Video Approximate value of function using tangent line Notes 2.5 Video Given graph of g (x), find when g'0, g'2.5 Video Given table of h' (x), find when h (x) is increasing /decreasing & tangent approximation. 2.5-2.9-2.99-2.999-2.9999; Use the information from (a) to estimate the slope of the tangent line to (f left( x right) ) at (x = - 3 ) and write down the equation of the tangent line. Show All Solutions Hide All Solutions.
- 2.5 Derivative As Tangent Lineup Calculus Formula
- 2.5 Derivative As Tangent Lineup Calculus Equation
- 2.5 Derivative As Tangent Lineup Calculus Solver
- 2.5 Derivative As Tangent Lineup Calculus Calculator
How do you find the equation of a line tangent to the function #y=x^2-5x+2# at x=3?
2 Answers
Explanation:
Let #y=f(x)=x^2-5x+2#
At #x=3,y=3^2-5*3+2#
#=9-15+2#
#=-6+2#
#=-4#
So, the coordinate is at #(3,-4)#.
We first need to find the slope of the tangent line at the point by differentiating #f(x)#, and plugging in #x=3# there.
#:.f'(x)=2x-5#
At #x=3#, #f'(x)=f'(3)=2*3-5#
#=6-5# Stock projectmr. regan's educational website builder.
#=1#
So, the slope of the tangent line there will be #1#.
Now, we use the point-slope formula to figure out the equation of the line, that is:
#y-y_0=m(x-x_0)#
where #m# is the slope of the line, #(x_0,y_0)# are the original coordinates.
2.5 Derivative As Tangent Lineup Calculus Formula
And so,
#y-(-4)=1(x-3)#
#y+4=x-3#
#y=x-3-4#
#y=x-7#
A graph shows us that it's true:
Explanation:
#y=x^2-5x+2#
#y' = 2x - 5#
2 Answers
Explanation:
Let #y=f(x)=x^2-5x+2#
At #x=3,y=3^2-5*3+2#
#=9-15+2#
#=-6+2#
#=-4#
So, the coordinate is at #(3,-4)#.
We first need to find the slope of the tangent line at the point by differentiating #f(x)#, and plugging in #x=3# there.
#:.f'(x)=2x-5#
At #x=3#, #f'(x)=f'(3)=2*3-5#
#=6-5# Stock projectmr. regan's educational website builder.
#=1#
So, the slope of the tangent line there will be #1#.
Now, we use the point-slope formula to figure out the equation of the line, that is:
#y-y_0=m(x-x_0)#
where #m# is the slope of the line, #(x_0,y_0)# are the original coordinates.
2.5 Derivative As Tangent Lineup Calculus Formula
And so,
#y-(-4)=1(x-3)#
#y+4=x-3#
#y=x-3-4#
#y=x-7#
A graph shows us that it's true:
Explanation:
#y=x^2-5x+2#
#y' = 2x - 5#
2.5 Derivative As Tangent Lineup Calculus Equation
At #x=3 :#
#y' = 2x - 5#
#y' = 6 - 5#
#y' = 1#
2.5 Derivative As Tangent Lineup Calculus Solver
#y = 3^2 - 5 xx 3 + 2#
#y = -4#
#y' = 1, (3, -4)#
#y - (-4) = 1(x - 3)#
#y = x - 7#