# How do you find an equation of the tangent line to the parabola #y=x^2-2x+7# at the point (3,10)?

The gradient of the tangent to a curve at any particular point is give by the derivative of the curve at that point. The normal is perpendicular to the tangent, so the product of their gradients is

so If

#dy/dx = 2x-2#

When

and

So the tangent we seek passes through

# y-10=4(x-3) #

# :. y-10=4x-12#

# :. y=4x-2#

We can confirm this graphically:

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To find the equation of the tangent line to the parabola y=x^2-2x+7 at the point (3,10), we need to find the slope of the tangent line at that point.

First, we find the derivative of the function y=x^2-2x+7 with respect to x.

The derivative of y=x^2-2x+7 is dy/dx = 2x-2.

Next, we substitute x=3 into the derivative to find the slope at the point (3,10).

dy/dx = 2(3)-2 = 4.

So, the slope of the tangent line at the point (3,10) is 4.

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 into the equation.

y-10=4(x-3).

Simplifying the equation, we get y-10=4x-12.

Finally, rearranging the equation, we find the equation of the tangent line to the parabola y=x^2-2x+7 at the point (3,10) is y=4x-2.

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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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