How do you graph the parabola #x^2 + 10x +8y +17 = 0# using vertex, intercepts and additional points?

1 Answer
Jul 21, 2018

Answer:

Vertex: #( -5,1)#, symmetry: #x=-5#, y intercept: #(0,-2.125)#, x intercepts: #(-7.83,0) and (-2.17,0)#, additional points : #( 3 ,-7) and -13,-7#

Explanation:

#x^2+10 x+8 y+17=0 or 8 y = -(x^2+10x ) -17# or

#y= -1/8(x^2+10 x ) -17/8 # or

#y= -1/8(x^2+10 x +25) +25/8-17/8 # or

#y= -1/8(x+5)^2 +1 # This is vertex form of equation ,

#y=a(x-h)^2+k ;(h,k)# is vertex , #h=-5 ,k=1,a=-1/8 #

Since #a# is negative, parabola opens downward.

Therefore vertex is at #(-5, 1)# Axis of symmetry is

#x= h or x = -5 ; # y-intercept is found by putting #x=0#

in the equation #0^2+10*0+8 y+17=0 or 8 y =-17# or

# y= -17/8= -2.125 # or at # (0,-2.125)#

x-intercepts are found by putting #y=0#

in the equation #x^2+10 x+17 = -8 y or x^2+10 x =-17# or

#x^2+10 x +25=25-17 or (x+5)^2= 8 or x+5= +- sqrt 8#

#:. x= -5+- sqrt 8 or x ~~ -7.83 , x ~~ -2.17 # , therefore ,

x-intercepts are at #(-7.83,0) and (-2.17,0)#

Additional points: # x= 3 , y=-1/8(3+5)^2 +1# or

# y= -1/8*8^2+1 or y = -7 :. ( 3 ,-7)#

# x= -13 , y=-1/8(-13+5)^2 +1# or

# y= -1/8*(-8)^2+1 or y = -7 :. ( -13 ,-7)#

graph{x^2+10x+8y+17=0 [-20, 20, -10, 10]}[Ans]