An object is launched upwards from an initial height of 4 feet above ground, with an initial velocity of 48 feet per second. If the object's position at time x is given by f(x) = −16x2 + vt + s, where v is initial velocity and s is initial height, which equation can be used to find the maximum height of the object after x seconds?

Answer:The equation [tex]f(x)=-16(x-1.5)^2+40[/tex] is used to find the maximum height.The maximum height of the object is 40 feet at x=1.5.Step-by-step explanation:The vertex from of a parabola is[tex]f(x)=a(x-h)^2+k[/tex]         ..... (1)where, a is constant, (h,k) is vertex.The object's position at time x is given by[tex]f(x)=-16x^2+vx+s[/tex]where, v is initial velocity and s is initial height.It is given that the initial height of the object is 4 feet and initial velocity is 48 feet per second.Substitute v=48 and s=4 in the above function.[tex]f(x)=-16x^2+48x+4[/tex]Rewrite the above equation in vertex form.[tex]f(x)=-16(x^2-3x)+4[/tex]If an expression is [tex]x^2-bx[/tex], then we need to add [tex](\frac{b}{2})^2[/tex] in it to make it perfect square.In the parenthesis b=3, [tex](\frac{3}{2})^2=(1.5)^2[/tex]Add and subtract (1.5)^2 in the parenthesis.[tex]f(x)=-16(x^2-3x+(1.5)^2-(1.5)^2)+4[/tex][tex]f(x)=-16(x^2-3x+(1.5)^2)-16(-(1.5)^2)+4[/tex][tex]f(x)=-16(x-1.5)^2+36+4[/tex]               [tex][\because (a-b)^2=a^2-2ab+b^2][/tex][tex]f(x)=-16(x-1.5)^2+40[/tex]            .... (2)The equation [tex]f(x)=-16(x-1.5)^2+40[/tex] is used to find the maximum height.On comparing (1) and (2), we geta=-16, h=1.5, k=40Therefore, the maximum height of the object is 40 feet at x=1.5.