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
Jun 18, 2016

Answer:

#y=-1/4x^2+3x-4#

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"#