How do you find the vertex and intercepts for #y = x^2 - 8x + 18#?

Question

Julia Adams · Accepted Answer

Vertex={(4,2)}, intercept={(0,18)}

#y=x^2-8x+18# #=>y'=2x-8# For #y'=0#, #2x-8=0# #=>x=4# #=>y=4^2-8*4+18# #=16-32+18# #=2# #=> vertex={(4,2)}# #Delta=(-8)^2-4*1*18# #=-8# #=> Delta<0# For #y=0#, #x^2-8x+18=0# #=># #x# is imaginary For #x=0#, #y=18# #=># intercept={(0,18)}

Avery Alexander · Answer

vertex: #(4,2)#
y-intercept: #18#
there is no x-intercept

Converting the given equation into vertex form: #y=m(x-a)^2+b# with vertex at #(a,b)# #y=x^2-8x+18# #rarr y=x^2-8x+16+2# #rarr y=1(x-4)^2+2# #color(white)("XXX")with vertex at #(4,2)# y-intercept is the value of #y# when #x=0# #rArr# y-intercept = 18# x-intercept is the value of #x# when #y=0# i.e. when #x^2-8x+18=0# but checking the discriminant (#b^2-4ac# using the standard form) we see that there are no solutions for #x# since the discriminant is #< 0#

Abigail Allen · Answer

undefined To find the vertex and intercepts for the quadratic function $ y = x^2 - 8x + 18 $, follow these steps:

1. **Vertex:** Use the formula $ x = -b / (2a) $ to find the x-coordinate of the vertex, where $ a = 1 $ (coefficient of $ x^2 $) and $ b = -8 $ (coefficient of $ x $). Substitute these values into the formula to get the x-coordinate of the vertex.

2. Once you have the x-coordinate of the vertex, substitute it back into the equation to find the y-coordinate of the vertex.

3. **Y-intercept:** Substitute $ x = 0 $ into the equation to find the y-intercept.

4. **X-intercepts:** Set $ y = 0 $ and solve the resulting quadratic equation for $ x $ to find the x-intercepts.

By following these steps, you can find the vertex, y-intercept, and x-intercepts of the quadratic function $ y = x^2 - 8x + 18 $.