The normal to the curve C when x=1 intersects the curve at point P. If C is given by f(x)=2x^2+5x-3, find the coordinates of P

Differentiate C:dy/dx=4x+5When x=1dy/dx=4(1)+5dy/dx=9This is gradient of tangent.Gradient of normal=-1/9When x=1, y=4y-4=-1/9(x-1)y=(-1/9)x+(37/9)(-1/9)x+(37/9)=2x^2+5x-30=2x^2+(46/9)x-(64/9)x=1 or x=-32/9P at x=-32/9y=365/81P(-32/9, 365/81)