find the point (x,y) at which the graph of y=8x^2+7x-3 has a horizontal tangent.
The function y= 8x^+7x-3 has a horizontal tangent at______ (ordered pair)
please and thank you, explanation would be greatly appreciated
The function y= 8x^+7x-3 has a horizontal tangent at______ (ordered pair)
please and thank you, explanation would be greatly appreciated
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A horizontal tangent will occur when the derivative of the function is zero.
df/dx = 16x + 7
0 = 16x + 7
x = -7/16 = -0.4375
Use this value of x to solve for y:
y = 8*(-0.4375)^2 + 7*(-0.4375) -3
= -4.53125
So, the horizontal tangent will occur at (-0.4375, -4.53125).
As an interesting (and not at all coincidental) side note, this is the minimum of f(x).
df/dx = 16x + 7
0 = 16x + 7
x = -7/16 = -0.4375
Use this value of x to solve for y:
y = 8*(-0.4375)^2 + 7*(-0.4375) -3
= -4.53125
So, the horizontal tangent will occur at (-0.4375, -4.53125).
As an interesting (and not at all coincidental) side note, this is the minimum of f(x).
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y = 8x^2 +7x-3 ...............(i)
Differentiate
dy/dx = 16x+7.........(ii)
For horizontal tangent m= tan(0 degrees) = 0
Therefore 16x+7 =0 or x =-7/16 ............(iii)
To find y, put the value of x in (i)
y = 8(-7/16)^2 +7(-7/16) -3
= 49/32 - 49/16 - 3= (49 -98 -96)/32 = -145/32
Hence the point is (-7/16, -145/32) ...................Ans
Differentiate
dy/dx = 16x+7.........(ii)
For horizontal tangent m= tan(0 degrees) = 0
Therefore 16x+7 =0 or x =-7/16 ............(iii)
To find y, put the value of x in (i)
y = 8(-7/16)^2 +7(-7/16) -3
= 49/32 - 49/16 - 3= (49 -98 -96)/32 = -145/32
Hence the point is (-7/16, -145/32) ...................Ans