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

Question

Answer

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

Answer

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.

Answer

Dy/dx = 2x -3 put x=2 dy/dx = 1i.e slope= 1therefore, =n of tangent is:y-y1 = dy/dx(x-x1)y-7 = 1(x-2)y-3 =x-2y= x-1orx-y-1=0

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