# How do you find the equation of tangent line to the curve #y=4x^3+12x^2+9x+7# at the given point (-3/2,7)?

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To find the equation of the tangent line to the curve y=4x^3+12x^2+9x+7 at the point (-3/2,7), we need to find the slope of the tangent line at that point.

First, we find the derivative of the given curve, which gives us the slope of the tangent line at any point on the curve.

The derivative of y=4x^3+12x^2+9x+7 is dy/dx = 12x^2+24x+9.

Next, we substitute the x-coordinate of the given point (-3/2) into the derivative to find the slope at that point.

dy/dx = 12(-3/2)^2+24(-3/2)+9 = 27/2.

So, the slope of the tangent line at the point (-3/2,7) is 27/2.

Using the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope, we can substitute the values to find the equation of the tangent line.

y - 7 = (27/2)(x - (-3/2)).

Simplifying this equation gives us the equation of the tangent line to the curve y=4x^3+12x^2+9x+7 at the point (-3/2,7).

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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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