How do you find the vertex and intercepts for #y = x^2 - 10x +17#?

1 Answer
Aug 18, 2017

vertex: #(5,-8)#

x-intercepts: #(5+2sqrt2,0),##(5-2sqrt2,0)#

y-intercept: #(0,17)#

Refer to the explanation for the process.

Explanation:

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]}