How do you identify the important parts of #y = (x – 1)^2 – 16# to graph it?

Question

Eli Assouline · Accepted Answer

Minimum point #rArr# #(1, -16)#
#y-#intercept #rArr# #(0, -15)# From the completed square, you already know that the minimum points are #(1, -16)# To find the #y-#intercept, you need to expand the brackets to get it in the form of #ax^2 + bx+c# , where #c# is your #y-#intercept. So, #(x-1)(x-1) -16# = #x^2-x-x+1-16# = #x^2-2x+1-16# = #x^2-2x+15# So #15# is your #y-#intercept. graph{(x-1)^2-16 [-9.495, 10.505, -17.76, -7.76]}

Abigail Allen · Answer

undefined To graph the function $ y = (x - 1)^2 - 16 $, you can identify the important parts as follows:

1. **Vertex:** The vertex of the parabola is at the point (1, -16).
   
2. **Axis of Symmetry:** The axis of symmetry is the vertical line passing through the vertex, which is $ x = 1 $.

3. **Direction of Opening:** Since the coefficient of $ x^2 $ is positive, the parabola opens upwards.

4. **Intercepts:** To find the x-intercept, set $ y = 0 $ and solve for $ x $. To find the y-intercept, set $ x = 0 $ and solve for $ y $.

5. **Symmetry:** The parabola is symmetric about its axis of symmetry.

By plotting these points and considering the direction of opening, you can sketch the graph of the function.