How do I find the vertex, axis of symmetry, y-intercept, x-intercept, domain and range of #y=x^2-10x+2#?

1 Answer
Nallasivam V
Oct 10, 2015

Vertex is #(5,-23)#
Axis of symmetry #x=5#
Y-intercept #y=2#
X- intercepts are #(9.8, 0) and (0.2, 0)#

Explanation:

Vertex

#3x=(-b)/(2a)=(-(-10))/(2 xx 1)=10/2=5#

At #x=5#

#y=5^2-10(5)+2#
#y=25-50+2#
#=27-50#
#y=-23#
Vertex is #(5,-23)#
Axis of symmetry #x=5#

Y-intercept

At #x=0#

#y=0^2-10(0)+2#
#y=2#

#b^2-(4ac)= (-10^2)-(4xx1xx2)#

X-intercept

At #y=0#

#x^2-10x+2=0#

#x=(-b+- sqrt(b^2-(4ac)))/(2a)#
#x=(-(-10)+- sqrt((-10)^2-(4 xx 1 xx 2)))/(2 xx 1)#
#x=(10+- sqrt(100-8))/(2)#
#x=(10+- sqrt(92))/(2)#
#x=(10+- 9.6)/(2)#
#x=(10+9.6)/2=19.6/2=9.8#
#x=(10-9.6)/2=0.4/2=0.2#

X- intercepts are #(9.8, 0) and (0.2, 0)#

graph{x^2-10x+2 [-58.5, 58.55, -29.25, 29.25]}

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