What are the vertex, focus and directrix of # y=-15+12x-2x^2 #?

Question

What are the vertex, focus and directrix of  # y=-15+12x-2x^2 #?

Eva Abramson · Accepted Answer

#(3,3),(3,23/8),y=25/8# #"since the equation has an "x^2" term, this is a"# #"vertically opening parabola"# #"the equation of a vertically opening parabola is"# #•color(white)(x)(x-h)^2=4a(y-k)# #"where "(h,k)" are the coordinates of the vertex and a"# #"is the distance from the vertex to the focus and directrix"# #"if "a>0" then opens upwards"# #"if "a< 0" then opens downwards"# #"to obtain this form "color(blue)"complete the square"# #y=-2(x^2-6x+15/2)# #color(white)(y)=-2(x^2+2(-3)x+9-9+15/2)# #color(white)(y)=-2(x-3)^2+3# #(x-3)^2=-1/2(y-3)# #4a=-1/2rArra=-1/8" parabola opens down"# #"vertex "=(3,3)# #"focus "=(h,a+k)=(3,23/8)# #"directrix is "y=-a+k=1/8+3=25/8# graph{(y+2x^2-12x+15)(y-0.001x-25/8)=0 [-10, 10, -5, 5]}

Eli Assouline · Answer

undefined To find the vertex, focus, and directrix of the parabola represented by the equation y = -15 + 12x - 2x^2, follow these steps:

1. Rewrite the equation in the standard form of a quadratic equation: y = ax^2 + bx + c.
2. Identify the coefficients a, b, and c from the equation.
3. Use the formula for finding the vertex of a parabola: 
   Vertex (h, k) = (-b/2a, f(-b/2a)), where f(x) represents the function.
4. The focus of the parabola can be found using the formula:
   Focus (h, k + 1/(4a)).
5. The directrix of the parabola is a horizontal line located at y = k - 1/(4a).

After finding the values of a, b, and c, substitute them into the formulas to find the vertex, focus, and directrix of the parabola.