Given:
#y=x^2-10x+17# is a quadratic equation in standard form:
#ax^2-10x+17#,
where:
#a=1#, #b=-10#, and #c=17#
Vertex: the maximum or minimum point of the parabola
axis of symmetry and #x# value for vertex: #x=(-b)/(2a)#
#x=(-(-10))/(2*1)#
#x=10/2#
#x=5#
To find the #y# value of the vertex, substitute #5# for #x# into the equation and solve for #y#.
#y=1xx5^2-10(5)+17#
#y=25-50+17#
#y=-8#
The vertex is #(5,-8)#. Since #a>0#, the vertex is the minimum point of the parabola, and the parabola opens upward.
X-Intercepts: the value of #x# when #y=0#
To find the x-intercepts, substitute #0# for #y# and solve for #x# using the quadratic formula.
#0=x^2-10x+17#
#x=(-b+-sqrt(b^2-4ac))/(2a)#
Plug in the known values.
#x=(-(-10)+-sqrt(-10^2-4*1*17))/(2*1)#
Simplify.
#x=(10+-sqrt(100-68))/2#
#x=(10+-sqrt32)/2#
Prime factorize #32#.
#x=(10+-sqrt(color(red)2xxcolor(red)2xxcolor(blue)2xxcolor(blue)2xx2))/2#
Simplify.
#x=(10+-color(red)2xxcolor(blue)2sqrt2)/2#
#x=(10+-4sqrt2)/2#
Simplify.
#x=(color(red)cancel(color(black)(10))^5+-color(red)cancel(color(black)(4))^2sqrt2)/color(red)cancel(color(black)(2))^1#
#x=5+-2sqrt2#
#x=5+2sqrt2,##5-2sqrt2#
x-intercepts: #(5+2sqrt2,0),##(5-2sqrt2,0)#
Y-Intercept: the value of #y# when #x=0#
To find the y-intercept, substitute #0# for #x# and solve for #y#.
#y=1xx0^2-10(0)+17#
#y=0+0+17#
#y=17#
y-intercept is #(0,17)#
Summary
axis of symmetry: #5#
vertex: #(5,-8)#
x-intercepts: #(5+2sqrt2,0),##(5-2sqrt2,0)#
y-intercept: #(0,17)#
graph{y=x^2-10x+17 [-15.68, 16.34, 5.38, 21.4]}
