Question 747753
The point P's coordinate -1 is the x coordinate.
Assuming it is, substitute -1 into the equation
y = 2x^2 + 6x + 7  to find the 'y' value.
y = 2(-1)^2 + 6(-1) + 7 
y = 2 + (-6) + 7
y = 3    So P's coords are (-1, 3)

Next differentiate 2x^2 + 6x + 7
              dy/dx = 4x + 6
Substitute x = -1 into the differentiated equation
                    = 4(-1) + 6 = 2 This is the gradient(m)
at point P.

   Using y - b = m(x - a)
Substitute P's x and y coords   (x = a and y = b)

         y - 3 = 2(x + 1)
         y - 3 = 2x + 2
             y = 2x + 2 + 3
             y = 2x + 5 This is the equation of the tangent at P
Hope this helps