given the dimensions of width and length of the picture are x+20 by 2x-10 and the frame is a constant 5 cm wider from the edge of the picture to the frame, than the area of the frame is defined as (x+30)(2x)-(x+20)(2x-10) = 30x+200.
If the width is equal to the length which I assume is true if the width is constant.
The y-intercepts are (0,0) and (5,0) The equation of the quadratic function is f(t) = at(t-5) Solving for a 12 = a(3)(3-5) a = -2 The equation of the quadractic function is f(t) = -2t(t -5) f(t) = -2t² + 10t<span />
