FOR WAT VALUE OF X IS THE TANGENT OF THE CURVE Y=X^2-4X+3 PARALLEL TO THE X AXIS
Note that the given equation is a parabola that opens upward. The only point on the graph where the tangent line is parallel to the x-axis is at the vertex of the parabola, where it is at minimum. The slope of the tangent line at this point is zero (y=0). To get this point, we get the derivative of y with respect to x:
y = x^2 - 4x + 3
0 = 2x - 4
x = 2 (answer)
if you want to get the point, we substitute this back to the original equation:
y = x^2 - 4x + 3
y = 2^2 - 4*2 + 3
y = -1
It occurs at (2, -1).
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To find the value of x for which the tangent of the curve y = x^2 - 4x + 3 is parallel to the x-axis, we need to determine when the slope of the tangent is equal to zero.
The slope of a curve can be found by taking the derivative of the equation with respect to x. So, let's find the derivative of y = x^2 - 4x + 3:
dy/dx = 2x - 4
To find when the slope is equal to zero, we set dy/dx equal to zero and solve the equation:
2x - 4 = 0
Adding 4 to both sides:
2x = 4
Dividing both sides by 2:
x = 2
Therefore, the tangent of the curve y = x^2 - 4x + 3 is parallel to the x-axis when x is equal to 2.