The curve C has the equation y = 2x^2 -11x + 13. Find the equation of the tangent to C at the point P (2, -1).
The first step is to differentiate the equation of the curve in order to find the gradient of the tangent at the curve. Remember that when differentiating polynomials, we multiply the index of the variable x, by its coefficient, then subtract 1 from the index. In addition, remember that x0 = 1.
In this case, dy/dx = 4x - 11.
Now if we plug in the x-coordinate of P (2) into dy/dx, we will get the gradient of the tangent to the curve at P.
dy/dx = 4(2) - 11
dy/dx = 8 - 11
dy/dx = -3.
Now we find the equation of the tangent using the formula for the equation of a straight line, and plugging in the coordinates of P:
y - y1 = m(x - x1)
y - (-1) = -3(x - 2)
y + 1 = -3x +6
3x + y - 5 = 0.
This is the equation of the tangent to the curve C at P.
JY
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