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The equation of the tangent to the curve f= x^2 - 3x + 2, at the point (2,7) is?
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The equation of the tangent to the curve f= x^2 - 3x + 2, at the point...
F '(x)=2x-3,put x=2 then the slope =1 so we have eq of tangent is (y-7)=1 (x-2)=x-y+5
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The equation of the tangent to the curve f= x^2 - 3x + 2, at the point...
Equation of Tangent to Curve


To find the equation of the tangent to the curve f= x^2 - 3x 2, at the point (2,7) we need to follow the given steps:


Step 1: Find the Derivative


The slope of the tangent is given by the derivative of the function f(x) at the point (2,7). Therefore, we need to find the derivative of x^2 - 3x + 2.


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


f'(x) = 2x - 3


Step 2: Find the Slope of the Tangent


Now, we can find the slope of the tangent at point (2,7) by substituting x = 2 in the derivative.


f'(2) = 2(2) - 3 = 1


Therefore, the slope of the tangent at point (2,7) is 1.


Step 3: Use Point-Slope Form to Find the Equation of the Tangent


The equation of the tangent at point (2,7) can be found using the point-slope form of a line.


y - y1 = m(x - x1)


Where (x1,y1) is the point on the line and m is the slope of the line.


Substituting the values of (x1,y1) = (2,7) and m = 1, we get:


y - 7 = 1(x - 2)


Therefore, the equation of the tangent to the curve f= x^2 - 3x + 2, at the point (2,7) is y = x + 5.


Step 4: Verify the Result


We can verify the result by graphing the curve and the tangent line. The graph confirms that the tangent line passes through the point (2,7) and touches the curve at that point.


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The equation of the tangent to the curve f= x^2 - 3x + 2, at the point (2,7) is?
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