# How do you find the equation of the tangent to the curve #y=x^2+2x-5# that is parallel to the line #y=4x-1#?

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To find the equation of the tangent to the curve y=x^2+2x-5 that is parallel to the line y=4x-1, we need to find the slope of the tangent line. The slope of the given line is 4.

To find the slope of the tangent line, we take the derivative of the curve y=x^2+2x-5. The derivative of y=x^2+2x-5 is dy/dx = 2x+2.

Since the tangent line is parallel to the given line, the slope of the tangent line is also 4.

Setting the derivative equal to 4, we have 2x+2=4. Solving for x, we get x=1.

Substituting x=1 into the original equation y=x^2+2x-5, we find y=1^2+2(1)-5 = -2.

Therefore, the point of tangency is (1, -2).

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

Substituting (1, -2) and m=4 into the point-slope form, we get y-(-2) = 4(x-1).

Simplifying, we have y+2 = 4x-4.

Rearranging the equation, we find the equation of the tangent line to be y = 4x-6.

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