The slope of a parabola 𝑦=3𝑥2−11𝑥+10y=3x 2 −11x+10 at a point 𝑃P is -5. Find the 𝑦−y− coordinate of the point 𝑃P.

The slope of a parabola given by the equation y = ax^2 + bx + c is given by the derivative of the function, which is 2ax + b.

In this case, the equation of the parabola is y = 3x^2 - 11x + 10. So, the derivative of the function is 2*3x - 11 = 6x - 11.

We are given that the slope of the parabola at point P is -5. So, we set the derivative equal to -5 and solve for x:

6x - 11 = -5 6x = 6 x = 1

Now that we have the x-coordinate of point P, we can find the y-coordinate by substitifying x = 1 into the equation of the parabola:

y = 3(1)^2 - 11(1) + 10 y = 3 - 11 + 10 y = 2

So, the y-coordinate of point P is 2.