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

Equation of normal to

Hence equation of normal given by point slope form will be

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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).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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