What is the equation of the line normal to #f(x)=-4x^2 -5x +1 # at #x=-3#?
The equation of the normal line is of the form:
So the equation is:
This is shown here:
graph{(-4x^2-5x+1-y)(-x/19-383/19-y)=0 [-80, 80, -40, 40]}
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To find the equation of the line normal to f(x) at x=-3, we need to find the derivative of f(x) and evaluate it at x=-3.
The derivative of f(x) is given by f'(x) = -8x - 5.
Evaluating f'(x) at x=-3, we get f'(-3) = -8(-3) - 5 = 24 - 5 = 19.
The slope of the line normal to f(x) at x=-3 is the negative reciprocal of f'(-3), which is -1/19.
Using the point-slope form of a line, where (x1, y1) is the point of tangency, we have y - f(-3) = -1/19(x - (-3)).
Simplifying this equation, we get y + 4 = -1/19(x + 3).
Therefore, the equation of the line normal to f(x) at x=-3 is y = -1/19x - 61/19.
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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.
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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