How do you write the equation of the parabola in vertex form given vertex (4,-1) and y intercept (0,15)?

2 Answers
Aug 30, 2017

See below.

Explanation:

Equation in vertex form is #a(x - h)^2 +k# where #h# is the axis of symmetry and #k# is maximum or minimum value.
In this example #h# is #- 4# and #k# is #- 1#:
So
#a(x + 4)^2 - 1# writing as a function: #y = a(x + 4)^2 - 1#

#y# intercept is #( 0 , 15 )#:

Hence: #y = a(x + 4)^2 - 1# => #15 = a( 0 + 4)^2 - 1 #
Solving for #a# gives #a = 15/16#

#15/16( x + 4)^2 - 1 #

Aug 30, 2017

#y=(x-4)^2-1#

Explanation:

#"the equation of a parabola in "color(blue)"vertex form"# is.

#color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))#
where (h , k ) are the coordinates of the vertex and a is a constant.

#"here " (h,k)=(4,-1)#

#rArry=a(x-4)^2-1#

#"to find a substitute "(0,15)" into the equation"#

#15=16a-1rArra=1#

#rArry=(x-4)^2-1larrcolor(red)" in vertex form"#
graph{(x-4)^2-1 [-10, 10, -5, 5]}