The point at which curve y=x^2-4x+3 has gradient -2?

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Answer 1 · 2018-02-11T20:08:32

#(1,0)#

#f(x) = x^2 - 4x+3#

power rule: #(Deltay)/(Deltax)(x^n) = nx^(n-1)#

#(Deltay)/(Deltax)(x^2) = 2x^1 = 2x#

#x^0 = 1#

#(Deltay)/(Deltax)(-4x^1) = -4x^0 = -4#

#(Deltay)/(Deltax)(3^0) = 0 * 3^(-1) = 0#

#(Deltay)/(Deltax)(x^2) + (Deltay)/(Deltax)(-4x) + (Deltay)/(Deltax)(3) = 2x - 4 +- 0#

#= 2x - 4#

#f'(x) = 2x - 4#

this is the gradient of curve at point #(x,y)#.

when the gradient is #-2#, #2x-4 = -2#

#2x = 2#

#x = 1#

to find the #y#-coordinate, use #f(x) = x^2 - 4x + 3#

#f(1) = 1 - 4 + 3#

#= -3 + 3#

#= 0#

the coordinates of the point are #(1,0)#

desmos.com/calculator

the tangent to the graph #f(x) = x^2-4x+3# at #(1,0)# is shown in blue.

its equation is #y = -2x + 2#.

#y = -2x + 2# is in #y = mx+c# form, so the gradient #m = -2#.

the gradient of the tangent to graph #f(x) = x^2-4x + 3# is #-2# at point #(1,0)#.

Answer 2 · 2018-02-11T20:19:09

#(1,0)#

#•color(white)(x)m_(color(red)"tangent")=dy/dx" at x = a"#

#dy/dx=2x-4=-2#

#rArr2x=2rArrx=1#

#x=1toy=1-4+3=0#

#rArrm=-2" at "(1,0)#