How do you write the standard form of the equation of the parabola that has the indicated vertex and whose graph passes through the given point: Vertex: (6, 5); point: (0, -4)?
1 Answer
Explanation:
The
#color(blue)"vertex form"# of the equation is.
#color(red)(|bar(ul(color(white)(a/a)color(black)(y=a(x-h)^2+k)color(white)(a/a)|)))#
where (h ,k) are the coordinates of the vertex.here the vertex = (6 ,5)
#rArry=a(x-6)^2+5" is the equation"# To find a , use the point (0 ,-4) that the parabola passes through.
Substitute x = 0 , y = -4 into the equation.
hence :
#a(0-6)^2+5=-4rArr36a=-9rArra=-1/4#
#y=-1/4(x-6)^2+5" is the vertex form of the parabola"# The
#color(blue)"standard form"# of the equation is.
#color(red)(|bar(ul(color(white)(a/a)color(black)(y=ax^2+bx+c)color(white)(a/a)|)))# To obtain this form expand and simplify the vertex form.
#-1/4(x-6)^2+5=-1/4(x^2-12x+36)+5#
#=-1/4x^2+3x-9+5=-1/4x^2+3x-4#
#rArry=-1/4x^2+3x-4" is the standard form"#
