How do you write standard form of quadratic equation of vertex (-1,5) and yint =3?

1 Answer
Jun 27, 2016

Answer:

#y=-2(x+1)^2+5#

Explanation:

The vertex form for a quadratic is
#color(white)("XXX")y=color(green)(m)(x-color(red)(a))^2+color(blue)(b)#
for a parabola with vertex at #(color(red)(a),color(blue)(b))#
#color(green)(m)# indicates if the parabola opens upward (if positive) or downward (if negative) and the "spread" of the parabola.

Since we want a parabola with vertex at #(color(red)(-1),color(blue)(5))#
It must have the form:
#color(white)("XXX")y=color(green)(m)(x+1)^2+5#

So what remains is to determine the value of #color(green)(m)#

W are told that #"yint" = 3# (I assume this means the y-intercept)

Remembering that the y-intercept is the value of #y# when #x=0#
by replacing #y# with #3# and #x# with #0# we have:
#color(white)("XXX")3=color(green)(m)(0+1)^2+5#

#color(white)("XXX")rarr 3=m+5#

#color(white)("XXX")rarr m=-2#

So the complete form for the equation of this parabola is
#color(white)("XXX")y=-2(x+1)^2+5#

Here is a graph of #y=-2(x+1)^2+5# to help verify our result:
graph{(-2)*(x+1)^2+5 [-6.735, 7.31, -0.513, 6.507]}