What is the equation of the line normal to #f(x)=4x^2-7x+6 # at #x=1#?

Question

What is the equation of the line normal to #f(x)=4x^2-7x+6  # at #x=1#?

Elijah Abram · Accepted Answer

Equation of normal to #f(x)# at #x=1# is #x+y=4#.

At #x=1#, #f(x)=4-7+6=3#, hence normal will be at point #(1,3)# on the curve #f(x)=4x^2-7x+6#. The slope of tangent will be given by value of #f'(x)# at #x=1#/ As #f'(x)=8x-7#, slope of tangent is #1#. As normal is perpendicular to tangent, slope of normal will be #-1/1=-1#; Hence equation of normal given by point slope form will be #(y-3)=-1*(x-1)# or #y-3=-x+1# or #x+y=4#.

Emilia Aguilar · Answer

undefined To find the equation of the line normal to f(x) at x=1, we need to determine the slope of the tangent line at x=1 and then find the negative reciprocal of that slope to obtain the slope of the normal line.

To find the slope of the tangent line at x=1, we can take the derivative of f(x) with respect to x and evaluate it at x=1.

The derivative of f(x) = 4x^2 - 7x + 6 is f'(x) = 8x - 7.

Evaluating f'(x) at x=1, we get f'(1) = 8(1) - 7 = 1.

Since the slope of the tangent line is 1, the slope of the normal line will be the negative reciprocal of 1, which is -1.

Now, we have the slope (-1) and the point (1, f(1)) on the normal line. We can use the point-slope form of a line to find the equation of the line.

Using the point-slope form, the equation of the line normal to f(x) at x=1 is y - f(1) = -1(x - 1).

Simplifying this equation, we get y - f(1) = -x + 1.

Therefore, the equation of the line normal to f(x) = 4x^2 - 7x + 6 at x=1 is y = -x + (f(1) + 1).